CHAPTER 1

1.1 Using the definition of acceleration:

α =∆v

∆t− 60 mph − 0

9.2 − 0=

88 ft/sec − 0

9.2 sec= 9.57 ft/sec2

1.2 Differentiate x(t) to obtain the velocity:

v(t) = x(t) = −10t + 88 ft/sec.

Differentiating again yields the acceleration:

a(t) = x(t) = −10 ft/sec2.

So v(t) = 0 = −10t + 88.

Solving this for t yields that:

v(t) = 0 at t = 8.8 sec.

1.3 Evaluating x at zero yields x(0) = 5 m.

Differentiating yields x(t) = v(t) = 3t2 − 2

so that v(0) = −2 m/s.

Likewise

a(t) = x(t) = 6t

so that a(0) = 0.

Now at t = 3 sec,

x(3) = 27 − 2(3) + 5 = 26 m,

v(3) = 27 − 2 = 25 m/s

and a(3) = 18 m/s2.

Since the velocity changes sign during this interval, the particle has doubledback and to compute the total distance traveled during the interval you mustcompute how far it travels before it changes direction and then add this to thedistance traveled after the particle has changed direction. The particle changesdirection when the velocity is zero, or at the value of t for which

v(t) = 3t2 − 2 = 0,

or at time

t = 0.8165.

The particle first moves from

x(0) = 5 m to x(0.8165) = 3.9 m or a distance of 1.1 m.

1

It then changes direction and moves from 3.9 m to

x(3) = 26 m. Thus it travels a total distance of

(26 − 3.9) + 1.1 = 1.1 + 1.1 + 21 = 23.2 m.

1.4 This again is straightforward differentiation:

v(t) = x(t) = 2t − 2.

This is zero when t satisfies:

2t − 2 = 0 or t = 1 sec.

Next:

a(t) = x(t) = v(t) = 2 ft/sec2 which is constant acceleration.

Alternately, the distance traveled can be computed directly from integratingthe absolute value of the velocity:

d =∫ 30 |3t2 − 2|dt = 23.178 m

1.5 Solution: v(t) = x(t) = 6 cos 2t m/s. a(t) = x(t) = v(t) = −12 sin(2t) m/sec2.Setting −12 sin(2t) = 0, yields 2t = π, or t = 0, π/2 s, π, 3π/2...nπ/2, for thetimes for the acceleration to hit zero.

1.6 From the definition, straightforward differentiation yields: xA = (3t2 + 6t) ft

so the vA = (6t + 6) ft/sec

and aA = 6 ft/sec2. Likewise, xB = 3t3 + 2t

so that vB = (9t2 + 2) ft/sec

and aB = 18t ft/sec2

a)Thus at t = 1 sec:

xA(1) = 9 ft, vA(1) = 12 ft/sec, and aA(1) = 6 ft/sec2

xB(1) = 5 ft, vB(1) = 11 ft/sec, and aB(1) = 18 ft/sec2

So that A is ahead of B and has the largest velocity, but B is accelerating fasterthe A.

b) Now at t = 2 sec:

xA(2) = 24 ft, vA(2) = 18 ft/sec, and aA(2) = 6 ft/sec2

xB(2) = 28 ft, vB(2) = 38 ft/sec, and aB(2) = 36 ft/sec2

Now B is ahead of A, and has larger velocity and acceleration

2

c) To find where A and B have moved the same distance, let xA(t) = xB(t) andsolve for t. This yields

3t2 + 6t = 3t3 + 2t, or t2 − t − 4/3 = 0 and t = 0

Solving yields t = +1.758s and t = −1.758s.

Here we are interested in the positive value of time so that

xA(1.758) = xB(1.758) = 3(1.758)2+6(1.758) = 3(1.759)3+2(1.758) = 19.81 ft.

1.7 Note that this is an inverse problem. Straightforward differeniation yields:

v(t) = 3(−e−t sin 10t + e−t10 cos 10t) = 3e−t(10 cos 10t − sin 10t)

a(t) = 3e−t(−102 sin 10t − 10 cos 10t) − 3e−t(10 cos 10t − sin 10t)

= −3e−t(99 sin 10t − 20 cos 10t)

1.8 From the definition

x(t) = t3 − 6t2 − 15t + 40 ft

so that:

x = v = 3t2 − 12t − 15 ft/sec,

and x = a = 6t − 12 ft/sec.

a) v(t) = 0 requires 3t2 − 12t − 15 = 0

or t2 − 4t − 5 = 0.

Solving for t yields:

t = −1 and 5.

Taking the positive value of t, the velocity is zero at t = 5 sec.

b) At t = 5 sec,

x(t) = x(5) = (5)3 − 6(5)2 − (15)(5) + 40 = −60 ft.

At rest t = 0,

x(0) = 40.

Then the particle has moved from 40 to -60 or 40 + 60 = 100 ft.

c) a(5) = (6)(5) − 12 = 18 ft/sec2.

1.9 Note: The purpose of this problem is to hit home the idea that the distancetraveled and the displacement are different. This problem is easiest to solveusing computational software as it involves plots. Students could also use asymbolic processor to compute the derivatives, although it would be a little over

3

kill and they must be reminded that simple derivations should be something theycan do in their head, on tests, while complicated derivatives are more accuratelydone with software. The plots are also easy to sketch by hand, but if they areinexperienced at plotting using software, it is best to start them off with somesimple plots.

a) v(t) = x = 0.6 cos 2t

a(t) = x = −1.2 sin 2t.

b) 3 sin 2t travels a distance of 0.3m in the time 0 to π4

sec and back another0.3m returning back to the origin from π

4to π

2s. So the total distance traveled

is 0.3 + 0.3 = 0.6m (maybe look at the plot first).

c) The position of the mass at t = π2s however is x

(

π2

)

= 0.3 sin(

2π2

)

= 0.

Note the position at any point is not always the distance traveled, which in b)is shown to be 0.6m.

t ..,0 .01π

2

x t .0.3 sin .2 tv t .0.6 cos .2 t

a t .1.2 sin .2 t

0 0.5 1 1.5 2

2

1

1

x t

v t

a t

t

FIGURE S1.9

4

1.10 Differentiation yields v(t) = 88.92 sin 0.26t ft, and

a(t) = 23.12 cos 0.26t ft/sec2.

The plot of each follows.

t ..,0 0.1 12.2

x t .342 1 cos .0.26 t v t .88.92 sin .0.26 t

a t .23.12 cos .0.26 t

0 2 4 6 8 10 12 14

500

500

1000

x t

v t

a t

t

FIGURE S1.10

1.11 The following code in Matlab computes the velocity from the displacement dataand plots it:

x=[8 9 11 13 14 15 17 18 22 27 32 37 41 44 46 48 49 49 48 47 46];

t=0;,01:02;

n=length (x);

v=0*x;

dt=.01;

for i=1:n-1

v(i+1)=(x(i+1)-x(i)/dt;

end

v

plot(t,v),xlabel(‘t*dt or elapsed time’), title (‘velocity versus time’)

5

This produces the following output:

v=

Columns 1 through 12

0 100 200 200 100 100 200 100 400 500 500 500

Columns 13 through 21

400 300 200 200 100 0 -100 -100 -100

And the following plot:

0 0.05 0.1 0.15 0.2 0.25-100

0

100

200

300

400

500

t*dt or elapsed time

velocity versus time

FIGURE S1.11a

The following produces the corresponding acceleration:

EDU>>a=0*v;

EDU>>for i=1:n-1

a(i+1)=(v(i+1)-v(i))/dt;

end

EDU>a

a=

Columns 1 through 6

0 10000 10000 0 -10000 0

Columns 7 through 12

10000 -10000 30000 10000 0 0

Columns 13 through 18

-10000 -10000 -10000 0 -10000 -10000

6

Columns 19 through 21

-10000 0 0

EDU>>plot(t,a),xlabel(‘elapsed time’), title (‘acceleration ersus time’)

0 0.05 0.1 0.15 0.2 0.25-1

-0.5

0

0.5

1

1.5

2

2.5

3x 10

4

elapsed time

acceleration versus time

FIGURE S1.11b

1.12 Consider the following Matlab code which uses the central difference to computethe velocity:

x=[8 9 11 13 14 15 17 18 22 27 32 37 41 44 46 48 49 49 48 47 46];

t=0:,.01:0.2;

n=length (x);

v=0*x;

dt=.01;

for i=1:n-1

v(i)=(x(i+1)-x(i-1))/(2*dt);

end

v

plot(t,v),xlabel(’t*dt or elapsed time’),title(velocity versus time’)

7

This results in the following values:

v =

Columns 1 through 12

0 150 200 150 0 100 150 150 250 0 450 500 500 450

Columns 13 through 21

350 250 200 150 50 -50 -100 -100 0

And the following plot:

0 0.05 0.1 0.15 0.2 0.25-100

0

100

200

300

400

500

t*dt or elapsed time

velocity versus time

The following is the Mathcad code for solving this problem:

velocity in mm/s versus time in s

. .1 19

i 1

xi 1 xi 1

.2 ∆t

0 0.05 0.1 0.15 0.2200

0

200

400

600

vi

t

i

v

t=0.01∆

8

1.13 Solution:

y(t) = −4.905t2 + 20t (m)

v = y(t) = −9.81t + 20 (m/s)

a = v = −9.81 (m/s)

The position, velocity and acceleration at t = 5 are

y(5) = −22.625 m, v(5) = −29.05 m/s, a(5) = −9.81 m/s2

Note that the ball returns to its initial state of y(5) = 0 when t satisfies−4.905t2 + 20t = 0 or, t = 20/4905 = 2.07 sec.

Then to obtain the total distance traveled by the ball, we need to calculatewhen the ball changes direction, i.e., when v(t) = 0.

v(t) = 0 = −9.81t + 20 or t = 2.0395.

From t = 0 to 2.039 sec. the ball travels a distance of y = 20.39 ft.

It then travels back past zero (the top of the building) another 20.39 ft. toy(0) = 0.

It then travels on a distance of y(5) = −22.625, beyond zero.

Thus the total distance traveled by the ball is 20.39 + 20.39 + 22.625 = 63.4 m.

1.14 Solution: x, v and a are given respectively by:

x(t) = e−ct sin ωt

v(t) = −ce−ct sin ωt + e−ctω cos ωt

a(t) = c2e−ct sin(ωt) − 2cωe−ct cos ωt − ω2e−ct sin ωt

x(0) = 0, v(0) = ω, a(0) = −2cω

1.15 Solution:

x(t) = 3t3 − 2t2 + 5, x(0) = 5m

v(t) = x(t) = 9t2 − 4t, v(0) = 0 m/s

a(t) = v(t) = 18t − 4, a(0) = −4 m/s2

1.16 The velocity and acceleration are respectively:

v(t) = 4 · t · cos(π · t) − 2 · t2 · sin(π · t) · πa(t) = 4 · cos(π · t) − 8 · t · sin(π · t) · π − 2 · t2 · cos(π · t) · π2

so that x(0) = v(0) = 0 and a(0) = 4

9

1.17 Solution:

y(t) = 3t2 − 20, so y(0) = −20 m

v(t) = y(t) = 6t, so v(0) = 0 m/s

a(t) = y(t) = 6, so a(0) = 6 m/s2

1.18 Solution:

x(t) = exp(−0.1 · t) · (3 cos t cos(2 · t) + sin(2 · t)), x(0) = 3

v(t) = 1.7 · exp(−.1 · t) · cos(2 · t) − 6.1 · exp(−.1 · t) · sin(2 · t), v(0) = 1.7

a(t) = −12.36exp(−0.1t) · cos(2t) − 2.79 · exp(−0.1t) · sin(2t), a(0) = −12.37

1.19 Solution:

x(t) = 5 · t − exp(−t) · 3 · t, x(0) = 0;

v(t) = 5 + 3 · exp(−t) · t − 3 · exp(−t), v(0) = 2;

a(t) = −3 · exp(−t) · t + 6 · exp(−t), a(0) = 6.

1.20 Solution: There is no derivative at t = 5, however the problem may be splitusing heaviside functions and differentiated over the two intervals.

t ..,0 .01 10

y t ..5 t Φ 5 t .25 .20 sin .π t π Φ t 5

y1 t .5 Φ 5 t ..20 cos .π t π Φ t 5 is the velocity

y11 t ...20 π2 sin .π t π Φ t 5 is the acceleration

0 5 10

200

200

y11 t

y1 t

t

FIGURE S1.20

10

During the 1st interval the slope dy/dt is constant. From the plot y(t) =20−04−0

t− = 5t for t = 0 to 5 sec. In the interval t > 5, y(t) = 25 + A sin(ωt + φ)where ω is the frequency and φ is the plane and A is the amplitude of the sinewave.

From the plot A = 20, the period T = 2 sec, so that ω = 2πT

= π. To find thephase evaluate y(5) = 25 = 25 + 20(sin(πt + φ) so that πt + φ = nπ, where n isany integer or φ = (n − 5)π, so φ = π will work. Thus:

y(t) ={

5t 0 < t < 520 sin(πt − π) t > 5

y′(t) ={

5 0 < t < 520π cos(πt − π) t > 5

y′′(t) ={

0 0 < t < 5−20π2 sin(πt − π) t > 5

1.21 Solution: aave = 60−0 mph5 sec

= 605

mphsec

= 125280 ftmile

milehour

1sec

1 hour3600 sec

= (12)(5280)300

ft/sec2

= 17.6 ft/sec2

17.6 ft/sec2 = 17.6 m3.25

ftsec2 = 5.4 m/sec2

5.4 m(sec)

9.81 m/sec2= 0.55 g’s or about 55% of a g.

1.22 One answer: a 4 min mile is near a record speed for trained runners so:

v = 1 mile4 min

= 14

milemin

1 min60 sec

· 5280 ftmile

= 22 ft/sec

or about 6.71 meter/s, or about 15 mph. On the other hand a sprinter cancover 100 m dash in 10 seconds, or 10 m/s.

1.23 The Matlab solution follows (see problem 1.11 and 1.12 also):

%first assign the data to the vector v

v = [0 0.2 0.27 0.3 0.3 0.3 0.3 0.3 0.35 0.4 0.5]; n=length(v);

%assign the time step

t=0:,0.1:11;

dt=0.1; a=0.*v; x=0.*v; % zeros in a and x

%using a loop (see statics supplement or student ed. of Matlab)

11

for i=1:n-1

x(i)=(v(i+1)+v(i))*(0.1)/2

end

for i=2:n

a(i)=(v(i-1)-v(i))/0.1

end

%to see the result plotted use the following

plot(t,x), xlabel (‘t*dt or elapsed time’), title (‘[position vs time’)

plot(t,a), xlabel (‘t*dt or elapsed time’), title (‘acceleration vs time’)

Note these plots are not given but appear in the text.

1.24 Solution:

Here is the Mathcad solution. From the plot estimate the following data:x0 0 x1 0 x2 0 x3 .02 x4 .04

x9 0.5 x10 0.7x5 .1 x6 0.18 x7 0.25 x8 0.38

These are measured every 0.1 sec. Thus the velocity becomes:n ..,0 1 10

x11 0

vn

x n 1 xn

0.1Which is plotted below

0 0.2 0.4 0.6 0.8 110

5

0

5

vn

.n 0.1

Next estimate the acceleration from the velocity:

which is plotted belowan

vn 1 vn

0.1

0 0.2 0.4 0.6 0.8 1100

0

100

an

.0.1 n

12

Next the equivalent Matlab code is given (without the plots as they look thesame). This is the same type of code used in problems 1.11, 1.12 and 1.23.

x=[0 0 0 0.02 0.04 0.1 0.18 0.25 0.38 0.7];

t=0:0.1:10; dt=0.1;

n=length(x);

v=0*x; a=0*v;

for i=1:n-1

v(i)=(x(i+1)-x(i))/0.1

a(i)=(v(i+1)-v(i))/0.1

end

plot (t,v), title (‘velocity versus time’)

plot (t,a), title (‘acceleration versus time’)

1.25 This is of second order and is linear x(t). The term sin(πt) is nonlinear but int, not x.

1.26 This is first order and linear in v(t).

1.27 This is first order and linear in v(t)

1.28 This is second order in θ and nonlinear because of the term sin θ(t).

1.29 This is second order in θ(t) and nonlinear because of the terms (dθ/dt)|dθ/dt|and sin θ(t).

1.30 This is first order in v and nonlinear because of the term v2.

1.31 This is second order and linear in x(t).

1.32 This is still linear of second order in x(t).

13

1.33 Solution:

a(t) = 3t2 − 4t, x(0) = 0, v(0) = 0

v(t) =∫ t0(3t

2 − 4t)t = t3 − 2t2 ft/sec

x(t) =∫ t0(t

3 − 2t2dt = 14t4 − 2

3t3 ft

The plot from Mathcad is:

t . .,0 0.01 2

0 1 22

1

0

x t

t

x(t): = 0.25 t2 -

2_3

t3

FIGURE S1.33

1.34 Solution:

a(t) = 40t cos πt∫ v3 dv =

∫ t0 40x cos πxdx

v(t) = 3 + 40π2 [cos πt + tπ sin πt − 1]

∫ x5 dx = 3t + 40

π2

∫ t0(cos πx + xπ sin πx − 1)dx

x(t) = 5 + 3t + 40π3 sin πt − 40

πt + 40

π3 [sin πx − πx cos πx]t0

x(t) = 5 + (3t − 40π

t) + 40π3 [2 sin πt − πt cos πx]

14

The following is typed in Mathcad to produce the desired plot

t ..,0 .01 10

x t 5 .340

π2

t .40

π3

.2 sin .π t ..π t cos π t

0 5 1050

0

50

x t

t

The equivalent Matlab code is:

sys t

x=5+(3-40/pin2)*t+(40/pin3)*(2*sin(pi*t)-pi*t*cos(pi*t))

ezplot(x,[0,10])

(the plot is suppressed here)

1.35 For the car t0 = 0, x0 = 0, vf = 60 mph = 88 ft/sec, tf = 6. For constantacceleration equation (1.28) yields at = vf or acar = 88

6= 14.67 ft/sec2. In

terms of g’s, a = 14.6732.2

= 46%g. For the sprinter t0 = 0, x0 = 0, vf = 10 m/sand xf = 15 m. From equation (1.30)

a =v2

f − v20

2(xf − x0)=

102 − 0

2(15 − 6)= 3.33 m/s2

In terms of g, a = 3.339.81

= 34%g.

1.36 a) This is constant acceleration with a0 = −9.81 m/s2 (taking positive x as up),v0 = 10 m/s and x0 = 0. Equation (1.30) relates the displacement, accelerationand velocity. At the top of the motion v = 0, so eq. (1.30) becomes

0 = 2(−9.81)(xtop) + v20 or xtop = 5.1 m

15

b) Equation (1.29) relates constant acceleration to velocity, time and position.When the ball returns to its initial position x = 0 and equation (1.29) becomes0 = a0t2

2+ v0t + 0. One solution is t = 0 which is the initial state. If t 6= 0, the

relationship becomes

t = v0

−2a0= 2.038 ∼ 2 sec

1.37 For constant acceleration v = at + v0 and v2 − v20 = 2ax so

a =v2−v2

0

2x= (5×106)2−(104)2

(2)(2)= 6.3 × 1012 m/s2,

t = v−v0

a= 5×106−1×104

6.3×1012 = 8.0 × 10−7 s.

1.38 Here a is a function of x, so consider the development of part 2, eq. (1.17) and(1.18)

adx = vdv or −kxdx = vdv or

−kx2

2|xx0

= v2

2|x0

kx20 = kx2 or

v2 = kx20 − kx2 or

v(x) =√

k(x20 − x2).

The position time relationship can be found from (1.20):

t =∫ t0 dt =

∫ xx0

dxv(x)

=∫ xx0

dx√k(x2

0−x2)

= 1√k

sin−1(

xx0

)

|xx0.

Rearranging and solving for x(t) yields

t =√

1k

sin−1(

xx0

)

|xx0or

√kt = sin−1

(

xx0

)

− sin−1(1) or√

kt+ π2

= sin−1(

xx0

)

or

x(t) = x0 sin(√

kt + π/2) = x0 cos(√

kt).

1.39 As the elevator starts from rest with constant acceleration to its operating speed

v2 = 2ax and v = at or

t = v/a = (3 m/s)/25 m/s2 = 1.2 sec

and travels a distance of

x = v2/2a = (3 m/s)2/2(2.5 m/s2) = 1.8 m.

Which is also the time and distance required to stop the elevator. Hence 2 ×1.2 s = 2.4 s and 2 × 1.8 m or 3.6 m are used up in starting up and slowingdown, the remaining distance 200 m - 3.6 m or 196.4 m is traveled at a constantvelocity of 3 m/s so

t = x/v = 196.4/3 m/s = 65.5 sec.

16

The total time is then

1.2 + 05.5 + 1.2 = 67.9 sec or a little over one minute.

1.40 Since a = −c2x case 2 applies and

v2 − v20

2=∫ x

x0

−c2xdx = −c2

(

x2 − x20

2

)

or v2 = v20 + c2x2

0 − c2x2. Substitution of x0 = 0, x = 10, v0 = 30, v = 0, yields0 = 302 + 0 − c2102 or c = 3.

1.41 Given a(x) = −cx2 x0 = 0, t0 = 0, v(0) = v0 we want to determine v(x). Fromeq. (1.17)

−c∫ x0 x2dx =

∫ vv0

vdv = 12(v2 − v2

0)

or12(v2 − v2

0) = − c3x3 or v(x) =

√

v20 − 2c

3x3

1.42 Since a is given as a function of velocity, case 3 applies: dx = vdv/f(v)

upon integrating

x − x0 =∫ vv0

vdv/(−v) = −v + v0.

Since x0 = 0 and v = 0 when it comes to rest, x = 750 mm.

1.43 From the problem statement y0 = 40 km, v0 = 6000 km/s calculate an expres-sion for y. Here acceleration is a function of position, so equations (1.17)-(1.20)apply. Given

a(y) = −g0R2

(R+y)2, g0 = 9.81 m/s2, R = 6370 × 103.

At t = 0, y0 = y(0) = 40 × 103 m, v0 = v(0) = 6000 m/s

Note ymax will occur when v = 0. So compute v.

a = dvdt

= dvdy

dydt

= −g0R2

(R+y)2. Integrating yields

∫ 0v0

vdv = −gR2∫ ymy0

dy(R+y)2

= g0R2[

1(R+ym)

− 1(R+y0)

]

, or

02

2− v2

0

2= g0R

2[

1R+ym

− 1R+y0

]

.

Thus60002

2= g0R

2[

1(R+y0)

− 1(R+ym)

]

or

4.552 × 10−8 =[

1R+y0

− 1R+ym

]

.

Solving for ym yields ym = 2575.

17

1.44 Solution:∫ 0vesc

vdv = −g0R2∫∞y0

dy(R+y)2

v2esc

2= −2g0R

2∫∞y0

dy(R+y)2

−v2esp = −2g0R

2∫∞y0

dy(R+y)2

= g0R2

limym → ∞

[

1R+ym

− 1R+y0

]

= −2g0R1

(R+y0).

That is −v2esp = −2g0R

2 1(R+y0)

Then: vesp = R√

2g0/(R + y0) = 11.14 km/s = 11.14 × 103 m/s

1.45 Given: a(v) = −cv = −0.4v, v0 = 100 km/hr.

Since a = dv/dt we have∫ vv0

dvv

= −ct or ℓn vv0

= − = ct.

Thus v = v0e−ct.

But v = dx/dt so that

x = x0 + v0

∫ t0 e−ctdt = x0 +

(

−v0

c

)

e−ct|t0Thus x(t) = x0 + v0

c(1 − e−ct).

With x0 = 0, v0 = 100 and c = 0.4 this becomes x(t) = 250(1 − c−0.4t).

1.46 Solution:

a(t) = 5 sin(20t) m/s2 x(0) = 1 m and v(0) = 3 m/s.

Integrating:∫ v0 dv = v − v0 =

∫ t0 5 sin 20αdα = − 5

20(cos 20t − 1)

where α is used as the “dummy” variable of integration. Then

v(t) = 3 − 0.25 cos 20t + 0.25 = 3.25 − 0.25 cos 20t m/s.

Integrating again yields

x(t) = x0 + 3.25t− 1.25 × 102 sin 20t

x(t) = 1 + 3.25t− 0.0125 sin 20t m

18

1.47 Solution:

v0 = 0.6 ft/s, a = −v3 ft/s2

Thus this is case 3 on page 19. However a straightforward integration of

a = dv/dt yields a(v) = −v3 = dvdt

.

Then dt = −dvv3 and integrating yields

t = − 12v2 + 1.389.

Rearrange to get v(t) =(

12t+2.778

)1/2ft/s.

At t = 4 sec, v(4) = 0.305 ft/s.

1.48 Follow example 1, because the acceleration is a function of velocity so case 3 isused.

Note that a = dvdt

= g − cv2 or dvg−cv2 = dt. Integrating both sides using the

stated initial conditions yields

t =∫ v

0

dv

g − cv2=

1

c

∫ v

v0

dvgc− v2

=(

1

c

)

1

2√

g/c

ln

√

g/c + v√

g/c − v

for gc

> v2 and 4(1c) 1

2√

g/cln

(

v−√

g/c

v+√

g/c

)

for v2 > g/c, from using a table of

integrals. Thus there are two possibilities. For g/c > v2; t = 1

2√

g/clm

[√g/c+v√g/c−v

]

.

Solving for v yields v(t) =√

gc

(

e2√

gct−1e2

√gct+1

)

; g/c > v2. For g/c < v2,

t = 12√

gcln

v−√

g/c

v+√

g/c

and solving for v(t) yields

v(t) =√

gc

(

1+e2√

gct

1−e2√

gct

)

; v2 > g/c.

Note from this second expression the v2 = g/c results in the expression −e2√

2ct =

e2√

2ct which has no solution. Thus v cannot reach the value g/c, i.e., v = g/c isthe driver’s terminal velocity. Next consider integrating again to calculate x(t),i.e., dx = vdt or

∫ x

0dx =

√

g

c

∫ t

0

(

e2√

gct − 1

e2√

gct + 1

)

dt

=

√

g

c

[

∫ t

0

e2√

gct

e2√

gct + 1dt −

∫ t

0

dt

e2√

gct + 1

]

x(t) =1

c

[

ℓn(

e2√

gct + 1)

−√gct − 0.693

]

19

1.49 This is a free fall problem or uniformly accelerated motion, where the accel-eration is given as g = 32.2 ft/s, and the time traveled can be determined byequation (1.29) with tf as the given. Equation (1.29) becomes

x(tf ) = 30 ft =gt2f2

+ v0tf + x0

Here v0 = 0, since the ball is dropped, x0 = 0 taking the window as the startingposition and hence

gt2f2

= 30 or tf =√

30/32.2 = 1.365 sec,

which is the time required to hit the ground. The expression for velocity underuniform or constant acceleration is equation (1.28) or (1.30). From (1.28)

v(tf) = a0(tf) + v0 or

v(1.365) = (32.2)(1.365) = 43.95 ft/sec.

1.50 This is a case of uniform acceleration a0 = g = 32.2 ft/s2, with v0 up, andtf = 1.71 s. Using eq. (1.29) again with v0 as the unknown yields

v(tf) = 30 − (g)t2f

2+ (−v0)tf + 0

Here −v0 is used because v0 is up and we have taken down as positive in writinga plus sign for a0(= g). This is consistent with the solution to 1.49. Solving forv0 yields

v0 =[

(g2)t2f − 30

]

/tf = 9.987 ft/sec.

1.51 Given a0 = 0.7g (constant acceleration), vf = 0 (because the car comes to astop). Convert mph to ft/s (60 mph = 88 ft/s, 45 mph = 66 ft/s, 30 mph = 44ft/s) and use eq. (1.30)

v2f = 2a0(xf − x0) + v2

0

where vf = 0, a0 = −0.7 g (minus because it decelerates), x0 = 0 (we start ourdistance measurement t = 0) then

xf =v2

0

−2a0=

v2

0

0.4g=

v2

0(ft/sec)2

(1.4)(32.2) ft/sec2

so that a) xf = 171.78 ft, b) xf = 96.63 ft, c) xf = 42.95 ft.

1.52 Following the solution to 1.52m with a0 = 0.4g yields

xp =v2

0

(−2)(−0.4)(32.2)

so that a) xf = 300.6 ft, b) xf = 169.1 ft, c) xf = 75.2 ft.

20

1.53 This is a uniform acceleration problem with a0 = 0.6g. Since the car starts fromrest v0 = 0 and asume x0 = 0. Let vf = 200 mph = 293.33 ft/sec. Then thetime to reach vf can be found from eq. (1.28)

vf = a0tf + v0 or tf = 293.33/(0.6)(32.2) = 15.183 s.

The distance traveled is found from eq. (1.29) to be (v0 = x0 = 0)

xf = a0t2f/2 = (0.6)(32.2)(15.183)2/2 = 2226.9 ft = 0.422 mile

1.54 Both cars undergo uniform acceleration aA = 0.9g and aB = 0.85g. Let themstart at t = 0 in the ame place from rest, i.e., xa(0) = xB(0) = vA(0) = vB(0) =0. Car A travels 1,000 m or takes the time determine by equation (1.29)

xA(tf ) = 1000 =(0.9g)(t2

f)

2

Then tf = 15.1 sec. During this time car B travels a distance determined by

xB(15.1) =aB(tf )2

2= (0.85)(9.81)(15.1)2

2= 950.6 m

So car A is 1000 - 950.6 = 49.4 m ahead of car B when it crosses the finish line.Note that if tf = 15.05 s is used and not rounded off, then the distance becomes55.47 mm insteady of 49.4 m.

1.55 This can be solved several ways including graphically by computing the areaunder the acceleration curve to generate the velocity, and the area under thevelocity versus time curve to compute the position:

First write the acceleration during each interval. For 0 < t < 50s,

a(t) = 2 m/s.

For 50 < t < 70s : a(t) = 0,

for 70 < t < 100, a(t) = 15(t − 70). Last for 100 > t > a(t) = 0.

Now calculate the area under the curve in each of these intervals being carefulto use the appropriate initial conditions at the beginning of each interval:

0 < t < 50 v(t) = 20t m/s

50 < t < 70 v(t) = 1000 m/s

70 < t < 100 v(t) = 7.5(t − 70)2 + 1000 m/s

100 > t v(t) = 7750 m/s so that v(120) = 7750 m/s

Integrating each of these in the interval yields

0 < t < 50 x(t) = 10t2

50 < t < 70 x(t) = 25000 + 1000(t − 50)

70 < t < 100 x(t) = 2.5(t − 70)3 + 1000(t − 70) + 45, 000

t > 100 x(t) = 7750(t− 100) + 142, 500

21

This last expression yields

x(120s) = 297, 500 m = 297.5 km.

1.56 The equation to be solved is of the formdvdt

+ v = et, v(0) = 0, x(0) = 1m

Comparing with equation (1.32) identifies p(t) = 1 and f(t) = et so that the

integrating factor becomes λ(t) = e∫

dt = et.

According to equation (1.34) the solution is then

v(t) = e−t(∫

etetdt + C) = e−t(12e2t + C)

at t = 0, v(0) = 0 ft/s so that C = 0.5 and v(t) = 12(et − e−t) m/sec

= sin h(t) m/sec

Integrating again yields the displacement

x(t) = x0 +∫ t0 e−τ (0.5e2τ − 0.5)dτ when x0 = 1 m. Thus x(t) = 1

2(et + e−t)m

= cos h(t) m

1.57 The equation to be solved is of the formdvdt

+ v = t, v(0) = 0, x(0) = 1, v(0) = 0

Comparing to equation (1.32): p(t) = 1 and f(t) = t. Thus the integrating

factor becomes λ(t) = e∫

dt = et

According to equation (1.34) the solution becomes

v(t) = e−t(∫

ettdt + C) = e−t[et(t − 1) + C] = t − 1 + Ce−t

At t = 0, v(0) = 0 so that 0 = −1 + C or C = 1. Thus: v(t) = t − 1 + e−t m/s

Integrating again yields (x0 = 1m)

x(t) = x0+∫ t0(t−1+e−t)dt = 1+ t2

2−t−e−t+1, so that x(t) = 2 − t + t2

2− e−tm

1.58 a) The equation to be solved is of the form

dvdt

+ tv = e−tt/2

Comparing this form to equation (1.3.2), identifies p(t) = t and f(t) = e−t2/2.Thus the integrating factor becomes

λ(t) = e∫

tdt = et2/2

Next, equation (1.34) yields that the solution is

v(t) = e−t2/2(∫

et2/2e−t2/2dt + c) = e−t2/2t + ce−t2/2

At 0, v(0) = 10 ft/s so that c = 10 and v(t) = 10e−t2/2 + e−t2/2t

22

Integrating again (x(0) = 0) yields x(t) =∫ t0(10 + τ)e−τ2/2dτ which yields the

error function when integrated, i.e., x(t) = 12.5 erf(0.71t) − et2/2 + 1.

b) This does not have an integrating factor, or other closed formed solution, sothe solution must be found numerically by writing the equation in first order forand applying an Euler or Runge-Kutta solution. A Mathcad solution is shown.

i ..0 4000 ∆t 0.001 ti.i ∆t

x0 1 v0 5

a ,v t .t2

v 1 .3 t

xi 1

vi 1

xi.vi ∆t

vi.a ,vi ti ∆t

0 1 2 3 4

5

10

15

xi

vi

ti

FIGURE S1.58

To solve this problem with Matlab, create and run the following code:

x(1)=1; v(1)=5; t(1)=0;

dt=0.001;

for n=1:4000;

x(n+1)=x(n)+v(n)*dt;

v(n+1)=(-t(n) .2*v(n)+1+3*t(n))*dt+v(n);

t(n+1)=t(n)+dt;

end

plot(t,x),plot(t,v)

23

1.59 First solve the homogeneous equation x + 5x + 4x = 0 by following eq. (1.41),assume a solution of the form x(t) = Aeλt where λ must satisfy

λ2 + 5λ + 4 = (λ + 4)(λ + 1) = 0

so that λ1,2 = −4,−1. Thus the homogeneous solution is of the form xh(t) =A1e

−4t +A2e−t. The particular solution is guessed to be xp = a+bt, of the form

of the forcing function where a and b are to be determined. Substitution of theassumed form for xp(t) into the equation of motion yields

5b + 4(a + bt) = 3t + 0to

Comparing coefficients of t and to yields

5b + 4a = 0 and 4b = 3

so that b = 3/4 and a = −54(3

4) = −15

16. Thus xp = −15

16+ 3

4t. This is called the

method of undetermined coefficients. The total solution is the sum (x = xh+xp)so that

x(t) = A1e−4t + A2e

−t + 34t − 15

16

To determine the coefficients A1 and A2 apply the initial conditions

x(0) = 0.5 = A1 + A2 − 1516

v(0) = 0 = −4A1 − A2 + 34

which represents two equations in the two unkowns A1 and A2. Solving yields

A1 = −0.229 and A2 = 1.667 and hence: x(t) = −0.229e−4t + 1.667e−t + 34t − 15

16

1.60 Define ω2 = km

= 41

= 4 and ζ = c2mω

= 2.795 > 0 so the system is over damped.Then the problem in standard form is

x + 2ζωx + ω2x = x + 5x + 4x = 0

Assume solutions of the form x = Aeλt. The characteristic equation becomesλ2 + 5λ + 4 = 0 which has roots λ1 = −1, λ2 = −4. Thus the general solutionis of the form

x(t) = A1e−t + A2e

−4t

Applying the initial condition yields

x(0) = 5 = A1 + A2

v(0) = 0 = −A1 − 4A2

which is a system of two linear equations in the two unknowns A1 and A2.Solving yields A1 = 20

3and A2 = −5

3. Thus the solution is x(t) = 20

3e−t − 5

3e−4t

and v(t) = 203(−e−t + e−4t)

24

1.61 The equation of motion has the form (after dividing by 1000)

a = dvdt

= −cv − 400x, v(0) = 0, and x(0) = 0.01m

Following along with equation (1.51), the Euler method of integration yields[

vi+1

xi+1

]

=[

vi − cvi∆t − 400xi∆txi + vi∆t

]

,[

v0

x0

]

=[

00.01

]

Using a high level language (Matlab, Mathcad or Mathematica) yields (somestudents may know the analytical solution for this equation. Others will knowhow to use the more sophisticated higher-order Runge-Kutta integration thefollowing plot. Values of c are varied until the plot produces only two oscilla-tions.)

c 15∆t .01

i ..01.5

∆t

v0

x0

0

.01

vi 1

xi 1

vi..c vi ∆t ..400 xi ∆t

xi.vi ∆t

0 0.5 1 1.50.005

0

0.005

0.01

xi

.i ∆t

FIGURE S1.61

Note here that the oscillation dies out at about t = 1 second, for a value of c =15, or a damping value of 15, 000 kg/s.

The Matlab code for doing this is given below using an Euler method. Thiscan also be done using ODE which involves a Runge-Kutta routine. Create the

25

following Matlab code then run it with different values of c until the desiredresponse results:

c=15

x(1)=0.01;v(1)=0.0;t(1)=0;

dt=0.01;

for n=1:150;

x(n+1)=x(n)+v(n)*dt;

v(n+1)=v(n)-c*v(n)*dt-400*x(n)*dt

end

plot(t,x)

Run this Matlab code with various values of c until the response decays withintwo cycles as desired.

1.62 Following the development of the numerical integration section equation (1.51)becomes[

vi+1

xi+1

]

=[

vi − 900vi∆t − 4000(xi)2∆t

xi + vi∆

]

with initial condition v0 = 0 and x0 = 20 mm. The Mathcad code is:

i ..0 1000

∆t .001

v0

x0

0

20

vi 1

xi 1

vi

..900 vi

∆t ...4000 xi

xi

∆t

xi

.vi

∆t

0 0.2 0.4 0.6 0.8 1

10

20

xi

.i ∆t

FIGURE S1.62

26

The equivalent Matlab code can be either Euler (see previous problem) orRunge-Kutta Method. To use RK, first save the following Matlab code un-der Onept62.m:

function xdot=onept62(t,x)

xdot=[x(2);-900*x(2)-4000*x(2)-4000*x(2)*x(2)];

% the equation of motion

Then the following commands will compute and plot the solution

EDU>tspan=[0 1] % defines the time interval of interest

EDU>x0=[20;0]; %enters the initial conditions, displacement first

EDU>ode45(*onept62’,tspan,x0); % calls the RK routine and applies

% it to 1.62.

1.63 Following the development of the numerical integration section (1.51) becomes

[

vi+1

xi+1

]

=[

vi − 90vi∆t − 100x3i ∆t

xi + vi∆t

]

with initial condition v0 = 0 and x0 = 10 mm.

The Mathcad code is:

∆t 0.001 N 500 i ..0 N ti.i ∆t c 90 k 100

v0 0 x0 10 a ,v x .c v .k x3

xi 1

vi 1

xi.vi ∆t

vi.a ,vi xi ∆t

0 0.1 0.2 0.3 0.4 0.55

0

5

10

xi

ti

FIGURE S1.63

The catch gets near zero within 1/20 second.

27

The Matlab code is to prepare the following file named onept63.m:

Function xdot=onept63(t,x)

c=20;k=100;

xdot=[x(2);-c*x(2)-k*x(1) 3];

Then type the following in the command window:

EDU>tspan-[0 0.5];

EDU>x0=[10;0];

EDU>ode45(‘onept63’,tspan,x0);

1.64 Following the solution to 1.63 equation (1.51) becomes:

[

vi+1

xi+1

]

=[

vi − cvi∆t − 100(xi)3∆t

xi + vi∆t

]

Repeat the numerical solution to problem 1.63 with successively smaller valuesof damping (c) each time until the solution oscillates twice before coming torest. A value of about c = 20 1/s comes close as illustrated.

The Mathcade code is:

∆t 0.001 N 500 i ..0 N ti.i ∆t c 20 k 100

v0 0 x0 10 a ,v x .c v .k x3

xi 1

vi 1

xi.vi ∆t

vi.a ,vi xi ∆t

0 0.1 0.2 0.3 0.4 0.510

0

10

xi

ti

FIGURE S1.64

28

The Matlab code requires the following file saved as onept64.m:

function xdot=onept64(t,x)

c=20;k=100; xdot=[x(2);-c*x(2)-k*x(1) 3];

The type the following in the command window

EDU>tspan=[0 0.5];

EDU>x0=[10;0];

EDU>ode45(‘onept64’,tspan,x0);

1.65 Solution: First set up the Euler form of the equation for numerical integration:

[

vi+1

xi+1

]

=[

vi − cvi|vi|∆t − 4kxi|xi|∆txi + vi∆t

]

Then resolve for various values of c, k and x0 until a response that dies out inone oscillation results. There are many answers, the plot shows this is achievedfor x0 = 0.01 m, k = 400 1/ms2, and c = 1000 m−1. Another solution is x0 = 2,c = 6 and k = 40.

The Mathcad solution is:

c 1000 ∆t 0.01 x0 .01v0 0

i ..0 3000

vi 1

xi 1

vi...c vi vi ∆t ...400 xi xi ∆t

xi.vi ∆t

0 5 10 15 20 25 30

0.005

0.005

0.01

xi

.i ∆t

FIGURE S1.65

29

Save the following Matlab code as a file named “onept65.m”:

function xdot=onept65(t,x)

c=1000;k=400;

xdot=[x(2);-c*x(2)*abs(x(2))-k*x(1)*abs(x(1))]

In the command window:

EDU>tspan=[0 30];

EDU>x0=[0.01; 0];

EDU>ode45(‘onept65’,tspan,x0);

1.66 Assuming r = xi + yj + zk so that mr = mxi + myj + mzk. Then mr = −gjyields x = 0, y = −g/m and z = 0. These are linear, decoupled equations.

1.67 Yields the 3 scalar equations x + cx = 0, y + cy + g = 0 and z + cz = 0 whichare decoupled, linear equations.

1.68 Assuming r = xi + uj + zk and v = vxi + vy j + vzk yields

x = −c√

(v2x + v2

y + v2z) vx, y = −c

√

v2x + v2

y + v2z (vy) and z = −c

√

v2x + v2

y + v2z

(vz). These are coupled, nonlinear equations.

1.69 Assuming r = xi + yj + zk and v = vxi + vy j + vzkj

x = 3t2, y = − sin(πt) and z = xz

The x and y equations are linear and decouple.

The z equation is nonlinear and coupled to x.

1.70 Consider the plane trajectory equations given by eq. (1.74). In this case weknow xf = 450 ft, zf = 12 ft, g = 32.2 ft/s2, x0 = 0, z0 = 0 and v0 = 130 ft/s.Thus equation 1.73 becomes

450 = (130 cos θ)t + 0

12 = −16.2t2 + (130 sin θ)t

which is two nonlinear algebraic equations in two unknowns t and θ. Solvingyields t = 6.86 sec, θ = 1.042 rad (59.7◦) and t = 4.088, θ = 0.561 rad (32.143◦).

30

These solutions are found using Mathcad. Note that there are two solutionsfound by taking different initial guesses to the iterative solution of these non-linear algebraic equations. One solution has a low angle which would drive intothe bunker and one (correct) that has a larger angle which will loft onto thegreen.

1.71 Choose (0,0) in the x − z plane to be on the ground so that vz(0) = 2 m/s,zf = 0, x0 = 0, z0 = 3m, xf = d, xf = 0. Then equation (1.73) becomesd = 2t, 0 = −9.81

2t2 + v0(0) + 3. Combining 4.905 t2 = 3 and t = d

2yields

d2 = 124.905

or d = 1.56 m.

1.72 Looing at the top half spray, Eq. 1.73 becomes

d2 cos 10◦ = 20 cos 70◦t + 0 (1)

d2 sin 10◦ = −16.1t2 + 20 sin 70◦t + 0 (2)

which is a system of two equations in the two unknowns: d2 and t.

From (1) t = d2 cos 10◦

20 cos 70◦= 0.1459d2 d2 = 20 sin 20◦(0.1439)−sin 10◦

(16.1)(0.1439)2or d2 = 7.588 ft.

Next consider the spray to the left:

d1 cos 10◦ = 20 cos 50◦t + 0 (1)

0 = −16.1t2 + 20 sin 50◦t + d1 sin 10◦ (2)

From (1) t = .0766d1 or t2 = .005268d21 and eq. (2) becomes

d1 = (20)2 cos2 50◦

(16.1)(cos210◦)(cos 10◦ tan 50◦ + sin 10◦) = 14.26 ft.

1.73 Working with equation 1.73 for projectile motion, let the hose be at x0 = z0 =0 and assume it hits at x(tf ) = x and z(tf ) = 0, then eq. (1.73) becomes

x = v0 cos θtf and 0 = gt2f2

+ v0 sin θtf . Solving this last expression for tf yieldstf = 2v0 sin θ

g, the time to hit the ground. Then from the expression for x

x(tf ) =2v2

0

gsin θ cos θ

The max value of x occurs at dx/dθ = 0 or2v2

0

g(− sin2 θ + cos2 θ) = 0. This

requires sin θ = cos θ or θ = 45◦, the value at which xf will be maximum.

31

1.74 From the problem statement, taking the batters “foot” as the origin, the valueof x0 = 0, z0 = 4 ft, v0 = 140 ft/s, θ = 20◦ and (since it hits the ground) zf = 0.Equation (1.73) then becomes

x(tf ) = 14 − cos 20◦ tf and 0 = −16.1t2f + 140 sin 20◦tf + 4

Solving the last expression for tf yields tf = 3.055 sec. and -0.081 sec. Obviouslythe physical value is tf = 3.055, which from the first equation yields xf =140(cos 20◦)(3.055) = 402 ft.

1.75 From the projectile equation for z: z = −16.1t2 +140 sin 20t+4. The maximumvalue of the parabolic trajectory would occur at tf/2 except the value of tfcalculated in 1.74 assumes the trajectory is 4 ft off the ground. The equationfor time of flight is −16.1t2f +140 sin 20◦tf = 0 or tf = 2.97 sec, and tf/2 = 1.487

sec. Then zmax = −16.1(

2.922

)2+ 140 sin 20

(

2.922

)

+ 4 = 39.6 ft.

1.76 Consider the projectile equation 1.73 and first solve for v0 so the ball just clearsthe bottom window. Picking a coordinate system 1m off the ground yields

x0 = z0 = 0, xf = v0 cos 30◦ tf , zf = 2m = −9.812

t2f + v0 sin θtf

where xf = 6.5m. This yields two equations in two unknowns:

6.5 = v0(.886)tf or tf = 7.5v0

Thus v0 = 12.56 m/s. With yf = 3m, this becomes v0 = 19.18 m/s so that hemust kick through with a speed: 12.56 < v0 < 19.18 m/s.

1.77 The initial velocity is given as v0 = 10 m/s, x0 = z0 = 0, zf = 2m (for smallestand 3m for largest). xf = 6.5m so the first equation of (1.73) becomes (let tdenote the time at which the ball reaches the window).

6.5 = 1 − cos θt or t = 0.65/ cos θ

The second projectile equation yields

2 = −(

9.812

) (

0.65cos θ

)2+ 10 sin θ

(

0.65cos θ

)

which can be solved numerically for θ = 40.865◦. Changing the value of zf = 3mand repeating yields θ = 55.58◦. Thus he needs to kick through at an angle be-tween 40.9◦ < θ < 55.6◦ to make it through the window with an initial velocityof 10 m/s. Each of these two equations have 2 solutions so it will also make itin for 59.44◦ < θ < 66.28◦ on the high lofty solution.

32

1.78 The given values are z0 = d/√

2, v0 = 25 m/s, θ = 0. Equation 1.73 becomes

xf = 25t = d√2

or t = d25

√2, 0 = −9.81

2t2 + d√

2, or d =

√2

29.81

(

d2

(625)(2)

)

, sod = 180.2 m.

1.79 Sample 1.14 gives the equation for a particle in projectile motion with windresistances. The equations are nonlinear and coupled and must be solved nu-merically. The initial conditions of x(0) = 0, vx(0) = 25, y(0) = 0, vy(0) = 0will allow the solution computed numerically following sample 1.14. The tra-jectory can then be plotted along with a line at 45◦ representing the hill. Theintersection will yield the value of d. Since we do not know d, it is best to putthe coordinate system at the end of the ski run and let z (or y) evolve in thenegative direction. Such a line passing through the origin has slope -1 and canbe written as d = −x, or di = −xi in incremental form. The Mathcad code is

i ..0 1100 ∆t 0.005 c 0.04 g 9.81

vx0 25 x0 0 vy0 0 y0 0

vxi 1

xi 1

vyi 1

yi 1

vxi..c vxi ∆t

xi.vxi ∆t

vyi.g .c vyi ∆t

yi.vyi ∆t

di xi

0 20 40 60 80 100 120 140

150

100

50di

yi

xi

FIGURE S1.79

Form the plot, then cross about 111 m out or d = 111/ cos 45◦ = 157 m downthe incline.

33

The Matlab code for solving and plotting is given next with the plots suppressedas they are the same as the above:

function xdot=onept78 (t,x):

c=0.04; g=9.81;

xdot=[x(2);-(c*x(2);x(4);-g-c*x(3)];

In the command window:

EDU>tspan=[0 140]

EDU>x0-[0;25;0;0];

EDU>[t,x]=ode45(‘onept78’,tspan,x0);

EDU>d=-x(:1);

EDU>plot(x(:,1),x(:,3),‘t‘,x(:,1),d,‘*’)

1.80 This is just a repeat of the previous problem with a more accurate nonlineardamping term. The Mathcad code is:

i ..0 1000 ∆t 0.005 c 0.002 g 9.81

vx0 25 x0 0 vy0 0 y0 0

vxi 1

xi 1

vyi 1

yi 1

vxi..c .vxi vxi

2 vyi2 ∆t

xi.vxi ∆t

vyi.g .c .vyi vxi

2 vyi2 ∆t

yi.vyi ∆t

di xi

0 20 40 60 80 100 120

150

100

50di

yi

xi

FIGURE S1.80

34

In this case the skier makes it about 108 meters out or so which is 108 =d cos(45◦) so that d = 152.7 m.

The Matlab code is the same as the previous problem except the “x dot” be-comes:

xdot=[x(2);-c*x(2)*sqrt(x(2) 2+x(4) 2);x(4);-g-c*x(2)*sqrt(x(2) 2 + x(4) 2)];

1.81 Solution: 300 yards = 900 ft so that xf = x(tf ) = 900. Given that the ball isat zero to start with x0 = y0 = 0, and hits the ground at yf = 0. With θ = 90◦

given, the trajectory equations are

900 = v0 cos 9◦t + 0 (1)

0 = −16.1t2 + v0 sin 9◦t + 0 (2)

Solve (1) for t and (2) for v0 to get v0 = 306 ft/s (about 209 mph).

1.82 Solution: 200 mph = 293.3 ft/s. Here x0 = y0 = 0, θ = 9◦. Then eq. (1.71)becomes

x = (293.3) cos 9◦t = 289.7t (1)

0 = −16.1t2 + 45.8t (2)

From (2) t = 2.849 sec so from (1) x = 825.6 ft

35

1.83 This follows directly from the solution of sample 1.14 with the following valuesand equations: x0 = y0 = 0, c = 0.05, vx0 = 293.3 cos 9◦ ft/sec and vy0 =293.3 sin 9◦ ft/s.

i ..0 600 ∆t 0.005 c 0.05 g 32.2vx0

.293.3 cos .9 deg

x0 0 vy0.293.3 sin .9 deg y0 0

vxi 1

xi 1

vyi 1

yi 1

vxi..c vxi ∆t

xi.vxi ∆t

vyi.g .c vyi ∆t

yi.vyi ∆t

0 200 400 600 800 1000

20

20

40

yi

xi

FIGURE S1.83

From the figure x = 758 ft (found by using the trace funciton in Mathcad).

The Matlab code is given in the Matlab supplement.

1.84 Again use the projectile equations of eq. (1.73). Here: x0 = 0, y0 = 0 (soyf = 3 ft), xf = 20 ft, θ = 45◦ and hence

20 = v01√2t or t = 20

√2

v0

3 = −16.1t2 + v0 · 1√2t

Solving yields v0 =√

(16.1)80017

= v0 = 27.53 ft/s.

36

1.85 Using the projectile motion equations with x0 = y0, xf = 20 ft, yf = 3 ft andv0 = 30 ft/sec, equations (1.73) becomes

20 = (30) cos θt (1)

3 = −16.1t2 + (30) sin θt (2)

Substitution of t = 23 cos tθ

from (1) into (2) yields the transindental equation

(3) = −16.1 49 cos2 θ

+ 20 tan θ

Solving for θ yields θ = 33.7◦ and 64.8◦. Either angle will “work”, however thelower angle gives a trajectory up through the bottom of basket whereas the64.8◦ solution gives the lofty shot and then goes through the top of the hoop.The following Mathcad code solves the problem:

Specify the known parameters.

v 0 30 x 0 0 y 0 7 x f 20 y f 10

g 32.2

Initial guess for time and angle

θ .30 deg t f 3

Given

x f..v 0 t f cos θ x 0

y f.g

t f2

2..v 0 t f sin θ y 0

=Find ,θ t f1.132

1.568=

1.132

deg64.859

Angle in degrees

A second solution can be found with a lower angle.

v 0 30 x 0 0 y 0 7 x f 20 y f 10

g 32.2

Initial guess for time and angle These values are assumed less.

θ .10 deg t f 1

Given

x f..v 0 t f cos θ x 0

y f.g

t f2

2..v 0 t f sin θ y 0

=Find ,θ t f0.588

0.801=

0.588

deg33.69

Angle in degrees

The second solution is not valid as the ball would hit the net from below.

37

1.86 Using the projectile motion equations with x0 = 0, y0 = 10, θ = 0, v0 = 120mph = 176 ft/s and yf = 0 (i.e., hits the ground) yields

xf = 176t (1)

0 = −16.1t2 + 10 (2)

From (2) t = 0.7885 sec and from (1) xf = 138.7 ft

1.87 This again uses the projectile motion equation. a)Let x0 = 0, xf = 6ft, y0 = 0,so yf = 15−4 = 11 ft, θ = 80◦ and the unknown is v0. Equation (1.73) becomes

xf = 6 = (v0 cos 80◦)t + 0 (1)

yf = 11 = (v0 sin 80◦)t − 16.1t2 + 0 (2)

Substitute t = 6v0 cos 80

from (1) into to (2) to get

11 = 6 tan 80◦ − 16.162

v20 cos2 80◦

(3)

Solving yields v0 = 28.9 f/s. b) Repeating (a) with xf = 16 eq. (3) becomes

11 = 16 tan 80◦ − 16.1162

v20 cos2 80◦

or v0 = 41.4 ft/s.

1.88 This is circular motion with R = 150 ft, at = 12 ft/s2. Compute the time t

at which an = 24 ft/s2. From Eq. (1.84), at = αr so that α = at

r= 12 ft/s2

140 ft=

0.08 rad/s2 a constant. dwdt

= 0.08 so that w − w0 = 0.08t or w = 0.08t + w0.From (1.84) an = rw2 = 25 = 150(w0 + 0.08ts)

2(1), where ts = time to slip.

Also at ts, a =√

a2t + a2

r =√

252 + 122 = 27.73 ft/s2 at slip. Solving (1), with

ω0 = 0 for ts yields ts = 10.08

(

25150

)1/2= 5.104 s.

1.89 This is a circular motion with r = 2 m and at(t) = 6 sin πt(m/s2). The particlestarts at rest so that θ(0) = ω(0) = 0. For circular motion v(t) =

∫ t0 atdt =

∫ t0 6 sin πt = 6

π(1 − cos πt). Also ar = v2

r= 1

2( 36

π2 )(1 − cosπt)2 = 18π2 (1 − cos π6)2.

From equation (1.81) taking the magnitude of a(t) yields

a(t) =√

a2t + a2

n =√

62 sin2 πt + [ 18π2 (1 − cos πt)2]2.

38

The plots follow:

t ..,0 0.001 2

v t .6

π1 cos .π t at t .6 sin .π t an t .18

π21 cos .π t 2

a t at t 2 an t 2

0 0.5 1 1.5 2

5

10

time s

vel

m/s

acc

el m

/s^2

a t

v t

t

FIGURE S1.89

The Matlab code for producing the plots is given in the following file:

syms t % declares t symbolic

v=(6/pi)*(1-cos(pi*t)); an=0.5v 2;

at=6*sin(pi*t); a=sqrt(at 2+an 2)

ezplot(a,[0,2]), ezplot(v,[0,2])

1.90 From the solution to 1.89 a(t) = (36 sin3 πt + 183

π4 (cos πt− 1)4)1/2. Find ts whena(t) = 5. The answer can be seen from the plot given in figure S1.89 or fromsolving

52 = 36 sin2 πts + 182

π4 (cos πts − 1)4

for ts which has 2 solutions in the interval of interest. From Mathcad they are:ts = 0.312 s, and 1.688s.

1.91 Given r = 200 m, v = 30 km/hrs = 301× 103 m

1hr

3600 sec·hr= 8.33 m/s. For circular

motion, equation (1.81) yields

an = v2

4= 8.33

200− 0.3469 m/s2 or an = 0.35 m/s2.

39

1.92 This is circular motion starting from rest so that θ(0) = ω(0) = 0, with r = 4m.

a) α(t) = 2t2 r/s2 so that dωdt

= 2t2 and ω − ω0 = 23t3

or w(t) = 23t3. Thus from

eq. 1.84 an = r(ω)2 = 4 · 49t6 and at = 4(2t2) = 8t2. At t = 2s, an(2) = 113.8,

at(2) = 32 so that a(t) =√

11382 + 322 = 118.2 m/s2.

b) From eq. (1.82) d2sdt2

= ra = 8t2 so that dv = 8t2dt or v − 0 = 83t3. Thus

ds = 83t3dt and s(2) = 8

3

∫ 20 t3dt and the total distance traveled is s = 10.67 m.

1.93 Solution: α(t) = 3t2 − 2t rad/s2 with ω0 = θ0 = 0. Integrating yields dωdt

=

3t2 − 2t or∫ 20 dω =

∫ 60 3t2 − 2t or ω(t) = t3 − t2 rad/s. Likewise dθ = (t3 − t2)dt

so that θ(t) =∫ t0(t

3 − t2)dt = t4

4− t3

3or θ(t) = 1

4t4 − 1

3t3 rad.

1.94 Solution: α(t) = t cos(πt) rad/s2, ω0 = 2 rad/s, θ0 = 30◦ = π6

rad. Thus dω(t) =t cos πtdt or upon integrating ω − 2 =

∫ t0 x cos πxdx = 1

π2 [cos πt + πt sin πt − 1]so that ω(t) = (2 − 1

π2 ) + 1π2 (cos πt + πt sin πt) rad/s. Integrating again yields

θ − π6

= 1π3 [2 sin πt − πt − πt cos πt + 2π3t]

and

θ(t) = π6

+ 2t + 1π3 (2 sin πt − πt − πt cos πt) rad

1.95 Solution: ω0 = 3 rad/s and α(ω) = −2ω2 rad/s2. From eq. (1.87) α = ω dωdt

=

−2ω2. Solving yields∫ ω3

dωω

= −2∫ θ0 dθ, or ℓnω

3= −2θ. ω = 3e−2θ so that

dθdt

= 3e−2θ or∫ θ0 e2θdθ =

∫ t0 3dt = 3t. Evaluating the other integral yields

12e2θ|θ0 = 3t or e2θ − 1 = 6t

and 2θ = ℓn(6t + 1) or θ(t) = 12ℓn(6t + 1) and θ(10) = 1

2ℓn(61) = 2.055 rad.

1.96 Solution: ω0 = 2 rad/s and α(ω, t) = −0.01ω+4t rad/s2. Then dωdt

= −0.01ω+4twhich is a first order differential equation of the form: ω + 0.01ω = 4t, ω0 = 2.Using the integrating factor x(t) = e0.01t, equation (1.34) yields the solutionω(t) = e−0.01t [C + 40, 000e0.01t(0.0t − 1)]

Since ω(0) = 2, C = −3998 and

ω(t) = −3998e−0.01t + 4000(0.01t− 1)

ω(t) = −4000 − 3998e−0.01t + 40t rad/s = [0.01t − 1 + e−0.01t](4 × 104) rad/s

Since ω = dθdt

, integrating this ω(t) yields θ(t).

θ(t) = 400 + [0.005t2 − t − 0.01e−0.01t](4 × 104) rad

where 3998 has to be rounded to 4000.

40

1.97 The equation of motion can be written as dωdt

= 2 sin θ − 0.4ω. Since ω = dθdt

,this can be written as a second order nonlinear equation in θ:

θ + 0.4θ − 2 sin θ = 0 θ(0) = π6

and θ(0) = 0

which can be solved by numerical integration as suggested in equation (1.93).That is[

ωn+1

θn+1

]

=[

ωn + (2 sin θn − 0.4ωn)∆tθn + ωn∆t

]

,[

ω0

θ0

]

=[

0π6

]

which is plotted in the following Mathcad file:

i ..0 500 ω0 0 θ0

π6∆t 0.01

ωi 1

θi 1

ωi..2 sin θi

.0.4 ωi ∆t

θi.ωi ∆t

0 1 2 3 4 5

2

4

6

θi

.i ∆t

FIGURE S1.97

The Matlab code is:

function xdot=onept97(t,x);

xdot=[x(2);2*sin(x(1))-0.4*x(2)];

command window:

EDU>tspan=[0 5];

EDU>x0-[pi/6;0];

EDU>ode45(‘onept97’, tspan, x0);

41

1.98 This is a second order, nonlinear differential equation in θ which can be solvednumerically by using the first order form suggested in equation (1.93). Fromthe given form of θ + 0.02θ|θ| − 3 cos θ = 0 θ(0) = π

6and θ(0) = 0. Equation

(1.93) becomes

[

ωn+1

θn+1

]

=[

ωn + (3 cos θn − 0.02ωn|ωn|]∆tθn + ωn∆t

]

which is plotted below in Mathcad:

i ..0 500 ω0 0 θ0

π6∆t 0.01

ωi 1

θi 1

ωi..3 cos θi

..0.02 ωi ωi ∆t

θi.ωi ∆t

0 1 2 3 4 5

1

2

3

θi

.i ∆t

FIGURE S1.98

In Matlab the code is:

function xdot=onept98(t,x);

xdot=[x(2);3*cos(x(1))-0.02*x(2)*abs(x/2))];

command window

EDU>tspan=[0 5];

EDU>x0=[pi/6;0];

EDU>ode(‘onept98’,tspan,x0);

42

1.99 Follow sample 1.18. Given r(t) = 3 cos ti + 3 sin tj + 4tk m, differentiationyields v(t) = −3 sin ti + 3 cos tj + 4k and a(t) = −3 cos ti − 3 sin tj so thatv(3) = i+ j+4k and a(3) = i− j. The unit tangent vector et is calculated fromeq. (1.97) to be

et(3) = v(3)|v(3)| = −0.085i − 0.594j + 0.8k

at(3) = (a · et)et = 0, an = a − at = a so that

en = an

|an| = a|a| = 0.99i − 0.141j

eb = et × en = 0.133i + 0.792j + 0.6k

1.100 From problem 1.99

v(t) = −3 sin ti + 3 cos tj + 4k and a(t) = −3 cos ti − 3 sin tj m/s2

Note magnitude of both v(t) and a(t) is constant.

1.101 From equation 1.100, ρ(t) = v2(t)|an(t)| where v2(t) is v(t) ·v(t) = 9 sin2 t+9 cos2 t+

16 = 9 + 16 = 25. Compute |an(t)|. From eq. (1.97) et = v

|v| = 15(−3 sin ti +

3 cos tj + 4k. Then from eq. (1.98)

at(t) = (a · et)et = 15[(−3 cos t)(−3 sin t) + (−3 sin t)(3 cos t) + (0)(4)]et = 0.

Now an = a − at = a − 0 = a. Thus |an(t)| = |a(t)| =√

32 cos2 t + 32 sin t = 3.Thus

ρ(t) = v2(t)|an(t)| = 1

3[25] = 8.33 m, a constant so that the motion is circular, moving

at a constant angular velocity.

1.102 Given r(t) = t2 i+3tj+10 sin tk m, successive differentiation yields the velocityand acceleration:

v(t) = r(t) = 2ti + 3j + 10 cos tk m/s

a(t) = r(t) = 2i − 10 sin tk m/s2

From eq. (1.100) the radius of curvature is ρ(t) = v2(t)|an|

Now v2 = v·v = 4t2+9+100 cos2 t. Following eq. (1.97) et(t) = v

|v| , at = (a·et)et,

an = a−at and ρ(t) = v2

|an| . Programming these formulations and evaluating ateach value of t yields

a) t = 1s, ρ(1) = 7.23 m

b) t = 3s, ρ(3) = 126.645 m

c) t = 5s, ρ(5) = 13.346 m

43

The solution in Mathcad is given in the following:

t 5

v

.2 t

3.10 cos t

a

2

0.10 sin t

etv

vat ..a et et

an a at

ρ.v v

an=ρ 13.346

The Matlab code is

t=5;v=[2*t;3;10*cos(t)];a=[2;0;-10*sin(t)];

et=v/norm(v);at=dot(a,et)*et;an=a-at;

pro=dot(v,v)/norm(an)

1.103 Solution: r(t) = 12(1 + t2) and θ = πt2 so that r(t) = t and r(t) = 1, θ = 2πt

and θ = 2π. From equation (1.103)

v = rer + rθeθ = ter + πt(1 + t2)eθ

a = (r − rθ2)er + (rθ + 2rθ)eθ = (1 − 2π2t2(1 + t2))er + (π(1 + t2) + 4t2π)eθ

To plot the motion define t : 0, 0.1...2, define r = 1+t2

2and θ = πt2. Then let

x = r cos θ and y = r sin θ which is plotted below.

44

The Mathcad code is:

t ..,0 0.01 2

r t1 t2

2θ t .π t2

x t r t 8 cos θ t y t .r t sin θ t

20 10 0 10 20

4

2

2

y t

x t

FIGURE S1.103

The Matlab code is:

EDU>t=linspace(0,2);

EDU>r=(1+t 2)/2;th=pi*t. 2;

EDU>x=r.*8.*cos(th); y=r.*sin(th);

EDU>plot(x,y)

1.104 The value of v(t) is always positive. Note that |v(t)| =√

t2 + π2t2(1 + t2)2 =

t√

1 + π2(1 + t2)2 = ds/dt so that

s − s0 =∫ 20

√

t2 + π2t2(1 + t2)2dt = 18.977 m.

1.105 Solution: r(t) = 2 so that r = r = 0 and there is circular motion. θ(t) = sin πtso that θ = π cos πt and θ = −π2 sin πt. From equations (1.103):

v(t) = rθeθ = 2π cos πteθ and

a(t) = −2π2 cos2 πter − 2π2 sin2 πteθ

Let x = r cos θ and y = r sin θ to plot the motion as illustrated below.

45

The Mathcad solution is:

t . .,0 0.01 2

r t 2

θ t sin .π t

0

30

6090

120

150

180

210

240270

300

330

2

1.5

1

0.5

0r t

θ t

FIGURE S1.105

The Matlab code is:

EDU>linspace(0,2);

EDU>r=2;th=sin(pi*t);

EDU>x=r*cos(th);y=5*sin(th);

EDU>plot(x,y)

1.106 Solution:dsdt

= |v(t)| =√

4π2 cos2 πt = |2π cos πt|s − s0 =

∫ 20 |2π cos πt|dt = 8 m

1.107 Since the particle starts from rest, r(0) = 0, θ(0) = 0 r(0) = 1, and θ(0) = 0.To determine v(t) and a(t) from eqs.(1.103) we need r(t), θ(t), r(t) and θ(t)which we can calculate by integrating r and θ

r(t) − r(0) =∫ t0 r(x)dx =

∫ t0 2e−xdx = 2(1 − e−t)

r(t) − 1 = 2∫ t0(1 − e−x)dx so that r(t) = 1 + 2(t + e−t − 1) = −1 + 2t + 2e−t

Likewise

θ(t) − 0 = π∫ t0 dx = πt and θ(t) − 0 = π

∫ t0 xdx = πt2

2

46

Now from eqs. (1.103)

v(t) = 2(1 − e−t)er + [(2e−t + 2t − 1)πt]eθ

a(t) = [2e−t − (2e−t + 2t − 1)π2t2]er + [(2e−t + 2t − 1)π + 2(2 − 2e−t)πt]

1.108 Let r(t) = 1 + 2(t + e−t − 1) and θ(t) = πt2

2. Then define x(t) = r(t) cos θ(t),

y(t) = r(t) sin θ(t) and plot (Mathcad solution)

t ..,0 0.01 2r t 1 .2 t e t 1 θ t

.π t2

2

x t .r t cos θ ty t .r t sin θ t

3 2 1 0 1 2 3 4

4

2

2

y t

x t

FIGURE S1.108

The distance traveled is∫ 20 |v(t)|dt = 14.438 m

The Matlab code is:

EDU>linspace(0,2);

EDU>r=1+2*(t-exp(-t)-1);th=(pi*t. 2)/2;

EDU>x=r.*cos(th);y=r.*sin(th);

EDU>plot(x,y)

1.109 Given ω = 2π rad/s and r(t) = r0 + ra sin 2πt, to determine the accelerationrequires expressions for r, θ, r and θ. Since ω = 2π, dθ = 2πdt and θ = θ0 +2πt,θ = 2π, and θ = 0.

Likewise r = 2πra cos 2πt and r = −4π2ra sin 2πt. From eq. 1.103, ar =r − rθ2 = −4π2ra sin 2πt − (r0 + ra sin 2πt)4π2 = −4π2(2ra sin 2πt + r0) m/s2

aθ = rθ − 2rθ = (2)(2πra cos 2πt)(2π) = 8π2ra cos 2πt m/sec2

47

1.110 Solution: a) Let r0 = 2, ra = 1.5 < r0 and θ0 = 0, r(t) = 2 + 1.5 sin 2πt,θ(0) = 0 so θ(t) = 2πt. For one revolution let t = 0, 0.01...1 sec, then letx(t) = r(t) cos θ(t) and y(t) = r(t) sin θ(t) which is plotted below using Mathcad

t ..,0 0.01 2r t 2 .1.5 sin ..2 π t θ t ..2 π t

0

30

6090

120

150

180

210

240270

300

330

3210

r t

θ t

FIGURE S1.110

The Matlab code for this plot is

EDU>linspace(0,2);

EDU>r=2+1.5*sin(2*pi*t);th=2*pi*t;

EDU>x=r.*cos(th);y=r.*sin(th),

EDU>plot(x,y)

b) The velocity is v(t) = rer + rθeθ so that:

|v(t)| = |√

r2 + r2θ2| = π√

1.5 cos2 2πt + (2 + 1.5 sin 2πt)2

so s =∫ 10 |v(t)|dt = 14.407 m

48

1.111 Solution: ar = 2t, aθ = cos(πz), r(0) = .5 m, θ(0) = 0. Since it starts fromrest θ(0) = r(0) = 0. From eq. (1.103) r − rθ2 = 2t, rθ + 2rθ = cos(πt)which is a system of 2 coupled 2nd order equations which are nonlinear andinhomogeneous. The initial conditions (4) are given above. To solve numericallyfollow sample 1.22 (use x = r, y = θ)

r = rθ2 + 2t x = r · y2 + 2tr = x

θ = −2 rθr

+ cos πtr

: y = −2xy/r + cos πtr

θ = y

The Euler formula becomes (tn+1 = tn + ∆t)

xn+1 = (rn · y2n + 2tn)∆t + xn

rn+1 = rn + xn∆tyn+1 = (−2xn · yn/rn + cos πtr

rn)∆t + yn

θn+1 = yn∆t + θn

x0

r0

y0

θ0

=

0.500

Once these are solved the polar coordinate r(t) and θ(t) are given by the digitalrecord for rn and θn.

49

1.112 The trajectory is obtained by plotting rn cos θn vs. rn sin θn from above. TheMathcad solution is:

Trajectory in meters

i . .0 2000 ∆t 0.001

a ,,,,v r ω θ t .2 t .r ω2

α ,,,,v r ω θ t .1

rcos .π t ..2 v ω

t .i ∆t

v

0r

0

ω0

θ

0

0.5

0

0

vi 1

ri 1

ωi 1

θi 1

.a ,,,,vi

ri

ωi

θi

ti

∆t vi

ri

.v

i

∆t

ωi

.α ,,,,vi

ri

ωi

θ

i

ti

∆t

θ .ωi

∆t

x .r cos θ yi

.ri

sin θ

0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

yi

xi

0

i

i i i i

FIGURE S1.112

50

The equivalent Matlab code is:

function xdot=onept112(t,x);

xdot=[x(2);x(2)*x(3) 2+2*t;x(4);-2*x(1)*x(3)/x(1)+(cos(pi*t))/x(2)];

In the command window:

EDU>tspan=[0 2];

EDU>x0=[0;5;0;0]

EDU>[t,x]=ode45(‘onept112’,tspan,x0);

EDU>xc=x(:,2).*cos(x(:,4));ys=x(:,2).*sin(x(:,4));

EDU>plot(xc,ys)

1.113 The acceleration components in polar coordinates are given in eq. (1.04) to be

ar = r − rθ2 and aθ = rθ + 2rθ.

We are given r and θ which we need to integrate to get r, r, θ and θ. Firstconsider θ = 0 so that θ = constant = 1.5 rad/s. Integrating again yieldsθ(t) = 1.5t. Then the above becomes simply

ar = r − (1.5)2r, aθ = 2(1.5)r = 3r

Integrating r = 3 − 0.01r requires the solution of

r + 0.01r = 3

Which is a second order differential equation with particular solution rP = 3t0.01

.The homogeneous equation is r + 0.01r = 0 which has solution r1 = A andr2 = Beλt. Substitution yields

λ2 + 0.01λ = 0 or λ = −0.01 so the homogeneous solution is

rH(t) = A + Be−0.01t

and the general solution is

r = rH + rP = A + Be−0.01t − 300t

To get A and B apply the initial condition r(0) = 0.4 and r(0) = 0

r(0) = A + B = 0.4, r(0) = 0.01B + 300 = 0 or B = −30, 000

A = 30, 000.4

Thus

r(t) = −30, 000.4 − 30, 000e−0.01t + 300t m

Thus r = 300(1 − e−0.01t), r = 3e−0.01t so that

ar = 3e−0.01t − 2.25(30000.4− 30000e−0.01t + 300t)

aθ = (3)(30000.4) − 90, 000e−0.01t + 900t

51

1.114 The trajectory is a part of x(t) = r cos θ versus y = r sin θ. The form of r(t)and θ(t) are given in the previous problem. Let t = 0, 0.1...4 and plot y vs. x(t)as given below from Mathcad

t ..,0 0.1 4

r t 30000.4 .30000 e.0.01 t .300 t

θ t .1.5 t

0

30

6090

120

150

180

210

240270

300

330

200010000

r t

θ t

FIGURE S1.114

The Matlab code is:

EDU>t=linspace(0,4);

EDU>th=1.5*t;r=3000.4-3000*exp(-0.01*t)+300*t;

EDU>x=r.*cos(th);y=r.*sin(th);

EDU>plot(x,y)

1.115 In order to determine the velocity and acceleration from eq. (1.103), r(t), r(t), r,θ, θ and θ are needed. Since θ = π/4 so that θ = 0 and θ(t) = (π/4)t (assumingθ(0) = 0). Then r(t) = r(θ(t)) = 100 + 60 cos πt

4so that r(t) = −15π sin πt

4and

r = −15π2

4cos πt

4.

52

Thus from eq. (103)

v(t) = (−15π sin πt4)er + (π

4)(100 + 60 cos πt

4)eθ

and

a(t) =[

−154π2 cos πt

4−(

π4

)2 (

100 + 60 cos πt4

)

]

er +[

−152π2 sin πt

4

]

eθ

a(t) = −(100 + 120 cos πt4)π2

16er − [15π2

2sin πt

4]eθ

1.116 Here we need to find r(t) from the drawing and knowledge of θ(t). θ(t) =π4

sin πt, so that θ = π2

4cos πt and θ = −π

43 sin πt = −π2θ(t). From the drawing

300 = r(t) cos θ(t) so that

r(t) = 300 sec θ(t) = 300 sec (π4

sin πt)

r(t) = 75π2 tan(π4

sin πt) sec (π4

sin πt) cos πt

r(t) = 300[

π2

16sin2 πt tan(π

4sin πt) + π4

16cos2 πt(tan2(π

4sin πt) − 1)

]

sec(π4

sin πt)

Then

v = [75π2 tan(π4

sin πt)sec(π4

sin πt) cos πt]er + [75π2sec(π4

sin πt) cos πt]eθ

The aceleration becomes (in terms of θ, θ and θ)

a = 300 sec θ[(2 tan2 θ)θ2 + tan θθ]er + 300 sec θ[θ + 2 tan θθ2]eθ

53

1.117 Let θ be the angle between r(t) and the 60 mm line. Let β(t) be the anglebetween the 100 mm radius and the end of r(t). From this triangle

r(t) = 60 cos θ + 100 cosβ

and from the law of sines: (1) 60 sin θ = 100 sinβ. Differentiation of the lawof sines yields (2) 60 cos θθ = 100 cosββ. These last two expressions can beused to remove the β dependence. Differentiating r(t) and using (1) and (2) toremove the β term yields

r = −60 sin θθ − 100 sinββ = −60 cos θ tan βθ

where we note that eq. (1) can be used to remove the β dependence in favor ofθ. Now θ = constant = π is given, so r becomes

r = −60π sin θ − 60π cos θ tanβ, where β = sin−1(0.6 sin θ). From (1)

r = −60π2 cos θ + 60π2 sin θ tanβ − 602π2 cos2 θ100 cos3 β

which can be verified symbolically using one of the codes. With r, r and rgiven, θ = π, so that θ = πt and θ = 0, then a can be determined from equation(104):

ar = r − rθ2 and aθ = 2rθ = 2πr.

The Mathcad code for plotting this follows:

a r t ddr t .r t π2

a θ t ..2 dr t π

0 0.5 1 1.5 23000

2000

1000

0

1000

2000Accelerations vs time

time s

a r t

a θ t

t

Ac

ce

lera

tio

n i

n m

/s2

FIGURE S1.117

54

1.118 From the problem statement α(t) = 0.5 cos t. Integrating yields ω(t) = 0.5 sin t+ω0 = 0.5 sin t since it starts from rest. Integrating again (θ0 = 0) yieldsθ(t) = −0.5 cos t + C = −0.5 cos t + 0.5. Now recall the formulation of theprevious problem β(t) = a sin(0.6 sin(θ(t)). Then using the formulas developedin problem 1.117, ar and aθ can be determined. These are illustrated in thefigure where the derivatives are confirmed by symbolic computations.

t ..,0 0.01 7 θ t .0.5 cos t 0.5 ω t .0.5 sin t α t .0.5 cos t

β t asin .0.6 sin θ t r t .60 cos θ t 100 cos β t

First symbolicaly calculate the time derivative of r(t) and define it by rd(t)

d

d t.60 cos θ t 100 cos β t

..60 sin θ td

d tθ t ..100 sin β t

d

d tβ t

rd t ..60 sin θ td

d tθ t ..100 sin β t

d

d tβ t

Next symbolically calculate the second deriviative and denote it by rdd(t)

d

d t..60 sin θ td

d tθ t ..100 sin β t

d

d tβ t

..60 cos θ td

d tθ t

2

..60 sin θ td

d

2

2tθ t ..100 cos β t

d

d tβ t

2

..100 sin β td

d

2

2tβ t

rdd t ..60 cos θ td

d tθ t

2

..60 sin θ td

d

2

2tθ t ..100 cos β t

d

d tβ t

2

.100 sin β t

Now define the acceleration components in terms of r and its derivatives, theta and omega:

ar t rdd t .r t ω t2

aθ t ..2 rd t ω t .r t α t

0 2 4 6 8

100

50

50

100

ar t

aθ t

t

. d

d

2

2tβ t

FIGURE S1.118

55

1.119 The acceleration and velocity in cylindrical coordinates are given in eq. (1.109)and (1.111) and require r, r, r, z, z, θ and θ. Here r = 1.5, θ = π and z =0.5 cos 2πt so that r = r = 0, θ = 0, θ = 0, z = −π sin 2πt and z = −2π2 cos 2πt.Thus

v = 1.5πeθ − π sin(2πt)ez m/s

a = −1.5π2er − 2π2 cos(2πt)ez m/s2

The Mathcad code for generating this plot is:

i . .0 40 t .0.1 i

x .1.5 cos .π ti

yi

1.5 sin .π ti z

i.0.5 cos ..2 π t

i

10

11

0

1

00.2

0.20.4

,,x y z

i

FIGURE S1.119

The Matlab code for forming this plot is given in the following file:

i=(0:1:40);

x=1.5*cos(pi*i*0.1);y=1.5sin(pi*i*0.1);z=0.5*cos(pi*i*0.1);

plot3(x,y,z)

56

1.120 The distance traveled is calculated from dsdt

= |v| or s = s0 +∫ t0 |v|dt. Assuming

the particle starts out s0 = 0, s =∫ 20

√

(1.5π)2 + (π sin 2πt)dt = 10.398 m.

The total distance traveled can also be calculated using software. For example,

the Mathcad code for computing this distance follows:

v t

0

.1.5 π

.π sin ..2 π t

=d0

2

tv t 10.398

t . .,0 0.01 2

FIGURE S1.120

1.121 Solution: r(t) = R − zhR, = R − h−0.1ht

hR = R − R + 0.1Rt = 0.1Rt so that

r = R10

and r = 0. θ(t) = 2πt so that θ = 2π and θ = 0.

z(t) = h(1 − 0.1t) so that z = −0.1h and z = 0.

Thus

v = 0.1Rer + (0.1Rt)(2π)eθ − 0.1hez = 0.1Rer + 0.2πRteθ − 0.1hez, and

a = (−0.1Rt)4π2er + (R5· 2π)eθ = −0.4Rπ2ter + 0.4πReθ.

The particle reaches the bottom of the cone when z(t) = 0 or h(1 − 0.1t) = 0or at t = 10 sec. Since θ = 2πt, as 10 sec have pased, then θ has gone around10 times before it reaches the bottom (2π10 = 20π or 10 complete cycles).

1.122 Solution: for R = 3m and h = 5 m, v(t) = 0.3er + 0.6teθ − 0.5ez so that

|v| =√

(0.3)2 + (0.6t)2 + (0.5)2 as it takes 10 sec to travel to the bottom

s =∫ 100 |v|dt =

∫ 100

√

(0.3)2 + (0.6t)2 + (1.5)2dt = 94.67 m

1.123 From eq. (1.111) the given form of the differential equations are er : 3 =r − rθ2 (1), eθ : 2rθ = 0.1 (2) and from ez : z = −2. Since the particle startsfrom rest at zero, all the initial conditions are zero and since the z coordinate isdecoupled it can be directly integrated to yield z = −2t and z = −t2. Equations(1) and (2) for r and θ on the other hand are two coupled nonlinear equationswhich must be solved numerically. They are

r = rθ2 + 3

θ = −2 rθr

+ 0.1r

57

Putting these in 1st order form or Euler integration yields

rrθθ

=

0100

,

rrθθ

=

rθy2 + 3dr

0.1/r − 2drdθ/rdθ

The Euler equations then becomes (follow sample 1.22) as illustrated in

rn+1

drn+1

θn+1

dθn+1

=

rn + drn · ∆t(rn · dθ2

n + 3) · ∆t + drn

θn + dθn · ∆tdθn +

(

0.1rn

− 2drn·dθn

rn

)

· ∆t

,

r0

dr0

θ0

dθ0

=

0000

Figure S1.123 shows the integration along with a plot of the first 3 s usingMathcad.

i ..0 3000

∆t 0.001

ti.∆t i

ar ,,,r dr θ dθ 3 .r dθ2

aθ ,,,r dr θ dθ .1

r0.1 ..2 dr dθ

r0 1 dr0 0 θ0

0 dθ0

0

dri 1

ri 1

dθi 1

θi 1

dri.ar ,,,ri dri θ

idθ

i∆t

ri.dri ∆t

dθi

.aθ ,,,ri dri θi

dθi

∆t

θi

.dθi

∆t

0

30

6090

120

150

180

210

240270

300

330

1050

ri

θi

FIGURE S1.123

58

The required Matlab code is as follows:

function xdot=onept123(t,x);

xdot=[x(2);3+x(1)*x(4) 2;x(4);(0.1-2*x(2)*x(4))/x(1)],

In the command window:

EDU>tspan[0 3];

EDU>0=[1;0;0;0];

EDU>[t,x]=ode(‘onept123’,tspan,x0);

xc=x(:,1).*cos(x(:,3));yc=x(:,1).*sin(x(:,3));

plot(xc,yc)

1.124 Since R(t) = 0.3 + 0.1t2, R(t) = 0.2t and R(t) = 0.2. Since θ(t) = π sin(πt),θ(t) = π2 cos πt and θ(t) = −π3 sin πt. Since φ(t) = π

2te−t, φ(t) = π

2(e−t − te−t)

and φ(t) = π2(−e−t + (−e−t + te−t)) = π

2e−t(t − 2). Now from equation (1.116)

v(t) = 0.2teR + π2(0.3 + 0.1t2)e−t(1 − t)eφ+[0.3 + 0.1t2][sin(πt

2e−t)](π2 cos πt)eθ

From Eq. (1.118)

a(t) = [0.2 − (0.3 + 0.1t2)(π2(1 − t)e−t)2 − (0.3 + 0.1t2)[sin(πt

2e−t)]2(π4 cos2 πt)]eR

+[(0.3 + 0.1t2)(π2(t − 2)e−t) + 2(0.2t)(π

2e−t(1 − t))

−(0.3 + 0.1t2)(π2 cos πt)2 sin(πt2e−t) cos(πt

2e−t)] eφ

+[(0.3 + 0.1t2)(−π3 sin πt) sin(πt2e−t) + 0.4t(π2 cos πt) sin(πt

2e−t)

+2(0.3 + 0.1t2)(π2e−t(1 − t))(π2 cos πt) cos(πt

2e−t]eθ

1.125 From Sample Problem 1.24 we have a = g sin φeφ, and we are given that R = 200mm, (Rθ)0 = 600 mm/s and that φ0 = π/2, θ0 = 0, g = 9810 mm/s2. Sinceh is constant (R = 200 ⇒ θ0 = 3), h = θ0 sin2 φ0 = 3 rad/s. Now fromthe geometry of spherical coordinates the relation to rectangular coordinates is(from fig. 1.21)

x(t) = R sin φ cos θ, y = R sin φ sin θ, z = R cos φ

Now to solve the problem we need to integrate (via Eulerian) the two coupledequations

θ = hsin2 φ

and φ = h2 cos φsin2 φ

+ gR

sin φ

subject to the initial condition φ0 = π2, θ0 = 0, φ0 = 0.

59

The Euler formulation and a plot of the motion is given in the following Mathcadcode:

i ..0 800 ∆t 0.001 ti.∆t i h 3 R 200 g 9810

aφ φ.h2 cos φ

sin φ 2.g

Rsin φ

dθ φ3

sin φ 2

dφ0 0 φ0

π2

θ0 0

dφi 1

φi 1

θi 1

dφi.aφ φi ∆t

φi.dφi ∆t

θi.dθ φi ∆t

xi..R sin φi cos θi yi

..R sin φi sin θi zi.R cos φi

0 100

100

0

100

200

100

0

,,x y z

FIGURE S1.125

60

The Matlab code is (using an Euler method):

x(1)=0;p(1)=pi/2,th(3)=0;h=3;R=200;g=9810

dt=0.001; for n=1:800;

x(n+1)=x(n)+(h 2*cos(p(n))/(sin(p(n)) 2 + (g/R)*sin(p(n))*dt;

p(n+1)=p(n)+x(n)*dt;

th(n+1)=th(n)+(3/sin(p(n)) 2)*dt;

end

xc=R*sin(p).*cos(p);yc=R*sin(p).*sin(p);zc=R*cos(p);

plot3(xc,yc,zc)

1.126 This is a repeat of problem 1.125 for the case that θ0 = 0. Since θ0 and theconstant of motion θ sin2 φ = h must hold for all θ, h must be zero and thekinematic equations become φ = g

Rsin φ and θ = 0 which are numerically

integrated in the following figure (Mathcad):

i ..0 800 ∆t 0.001 ti.∆t i h 3 R 200 g 9810

aφ φ .g

Rsin φdθ φ 0

dφ0

0 φ0

π

2θ

00

dφi 1

φi 1

θi 1

dφi

.aφ φi

∆t

φi

.dφi

∆t

θi

.dθ φi

∆t

xi..R sin φ

icos θ

iyi

..R sin φi

sin θi

zi.R cos φ

i

00.20.40.60.81

100

0

100

200

100

0

,,x y z

FIGURE S1.126

61

The Matlab code is contained in the following file which when run will producean identical plot:

x(1)=0;p(1)=pi/2;th(1)=0;h=3;R=200;g=9810;

dt=0.001

for n=1:800;

x(n+1)=x(n)+(g/R)*sin(p(n))*dt;

p(n+1)=p(n)+x(n)*dt;

th(n+1)=th(n);

end

xc=R*sin(p).*cos(p);yc=R*sin(p).*sin(p);ac=R*cos(p);

plot3(xc,yc,zc)

1.127 From the solution to problem 1.126, φ = gR

sin φ for the case that the initialvelocity in the circumferential direction is zero (h = 0). We are given thatφ = π +β, β small so that using the trig identity for sin(π +β) we have sin φ =sin(π + β) = sin π cos β + cos π sin β = − sin β. But since β is assumed small,we can use the small angle approximation: − sin β = −β, and our differentialequation becomes (φ = d2

dt2(π + β) = β)

β + gRβ = 0

which was solved analytically in Sample Problem 1.9. The solution is

β(t) = β0 cos(√

gRt) +

√

gRβ0 sin(

√

gRt)

where β0 and β0 are the initial angle and angular velocity given to the particle.

1.128 From the value of a in spherical coordinates, the following 3 equations result

12 = R − Rφ2 − Rθ2 sin2 φ

4 = Rφ + 2Rφ − Rθ2 sin φ cosφ

5 = Rθ sin φ + 2Rθ sin φ + 2Rφθ cos φ

which can be numerically integrated subject to the initial conditions

R(0) = 2θ(0) = 0

φ(0) = π/2

R(0) = 2θ(0) = −2φ(0) = 1/2

(since vθ = −4 = R(0) sinφ(0)θ(0))(since vφ = R(0)θ(0) + 2θ(0) + 1)

by direct comparison with v(0) = 2er + eφ − 4eθ.

62

Once one numerically integrated, the expressions x = R sin φ cos θ, y = R sin φ sin θand z = R cos φ can be used to construct a 3-D plot of the motion as describedin the figure that follows using Mathcad.

i ..0 1000 ∆t 0.001 ti.∆t i

aR ,,,R vφ vθ φ 12 .R vφ2 ..R vθ2 sin φ 2

aφ ,,,,R vR vφ vθ φ .1

R4 ..2 vR vφ ...R vθ2 sin φ cos φ

aθ ,,,,R vR vφ vθ φ .1.R sin φ

5 ...2 vR vφ sin φ ....2 R vφ vθ cos φ

vR0 2 R0 2 vφ0 0.5 φ0

π2

vθ0 2 θ0 0

vRi 1

Ri 1

vφi 1

φi 1

vθi 1

θi 1

vRi.aR ,,,Ri vφi vθi φi ∆t

Ri.vRi ∆t

vφi.aφ ,,,,Ri vRi vφi vθi φi ∆t

φi.vφi ∆t

vθi.aθ ,,,,Ri vRi vφi vθi φi ∆t

θi.vθi ∆t

xi..Ri sin φi cos θi yi

..Ri sin φi sin θi zi.Ri cos φi

1510

50

5

0

3

2

1

0

,,x y z

FIGURE S1.128

63

The required Matlab code is:

R(1)=2;vR(1)=2;p(1)=pi/2);vp(1)=0.5;th(1)=0;vth(1)=-2;

dt=0.001

for n=1:1000;

R(n+1)=R(n)+vR(n)*dt

vp(nt)=vp(n)+((1/R(n))*4-2*vR(n)*vp(n)+R*vth(n) 2*sin(p(n)*cps)xdt;

p(n+1)=p(n)+vp(n)*dt;

vth(n+1)=vth(n+1)+(1/(R(n)*sin(p(n)))*(5-2*vR(n)*vp(n)*sin(p(n))

-2*R(n)*vp(n)*vth(n)*cos(p(n))*dt;

th(n+1)=th(n)+vth(n)*dt;

vR(n+1)=vR(n)+(12+R(n)*vp(n) 2+R(n)*vth(n) 2 * sin *p(n))+ dt;

end

x=R.*cos(p).*sin(th);y=R.*sin(p).*sin(th);z=R.*cos (th);

plot3(x,t,z)

1.129 We are given xA = 2t + 6 and xB/A = 3t2 + 2t + 3. From equation (1.120)

xB = xB/A + xA = 3t2 + 4t + 9 so that xB = vB = (6t + 4) m/s, and aB = 6 m/s2.

Likewise vB/A = 6t+2 and aB/A = 6 m/s2 from straight forward differentiation.

1.130 Since the particles start from rest vA(0) = vB(0) = 0 and since vB/A(0) =vB − vA, vB/A(0) = 0. Since xA = 3t + 2 and xB = 6t2 + 2, vA = 3 and aA = 0,vB = 12t and aB = 12. Since aB and aA are both constant and different in value:the two particles never have the same acceleration. The two particles reach thesame position when xB/A = 0. xB/A = 6t2 + 2 − (3t + 2) = 6t2 − 3t = 0 whent = 0 and when t = 1/2s. Since xB/A = 6t2 − 3t, vB/A = 12t − 3 so that theyhave the same velocity when t = 3/12 or t = 1/4s.

1.131 Here you must pick t0 carefully as the cars do not start moving at the same time.Car B moves with a constant velocity of vB = 15 m/s so that xB = 15t+xB(0)for all time. Car A on the other hand, starts moving when car B crosses aposition 100 meters out. Starting the clock at 0 when car B crosses the 100m mark and taking the position x = 0 m as the reference point we can writexB = 15t+100, since vB = 15 m/s implies that xB(t) = 15t+xB(0) = 15t+100.Now consider car A. We have that xA(t) = t2 until it reaches the speed of 20m/s. This happens when 2t = 20 or at t = 10 sec. Note that xA(10) = 100 m,

64

so that the cars cannot meet until after t > 10 s (i.e., car B started 100 m out).For t > 10 s, we are given that vA = 20 m/s. Integrating yields∫ xA

xA(10) dx =∫ t10 20dt or xA(t) = 20t − 200 + xA(10)

But xA(10) = 100. Then for t > 10, xA(t) = 20t− 100. Now for t > 10, xB/A =15t + 100− (20t− 100) = −5t + 200. Then xB/A = 0 yields tm = 40s, the timeat which the cars meet. They have traveled at distance of xA(40) = (20)(40)−100 = 700 m which can be checked by calculating xB(40) = (15)(40)+100 = 700m.

1.132 Particle B reverses direction when the velocity is zero. Since xB = 12 + 18t −4.9t2, vB = 18 − 9.8t = 0 at tr = 18

9.8= 1.83s, at this time particle B changes

directions. The two particles will meet when xA = xB or 5+2t = 12+18t−4.9t2,which has solution tc = −0.391, and 3.656s. So they collide at tc = 3.656s. Withrelative velocity vB/A(tc) = 16 − 9.8t = −19.83 m/s. since tc > tr, the particlescollide after B reverses direction as illustrated in the figure.

t ..,0 0.1 4

xB t 12 .18 t .4.9 t2 xA t 5 .2 t

0 1 2 3 4

10

20

30

xA t

xB t

t

FIGURE S1.132

1.133 Let car A be moving to the right with vA(0) = 60 mph = 88 ft/s, xA(0) = 0, andaA = −16 ft/s2. Let car B be moving to the left with xB(0) = 300, vB(0) = −88ft/s and aB = 15 ft/s2. Integrating each yields

vA = vA(0) − ∫ t0 16dt = 88 − 16t and

xA = xA(0) + 88t − 8t2 = 88t − 8t2

vB = vB(0) + 16∫ t0 dt = −88 + 16t

xB = xB(0) − 88t + 8t2 = 300 − 88t + 8t2

65

If they collide xB = xA or 300− 88t + 8t2 = −8t2 + 88t or t = 2.109 and 8.891.Now check to see if the cars are still moving at that time. The cars come torest at vA = vB = 0 or 88 − 16t = 0 or t = 88

16= 5.5 sec. Thus the cars hit at

t = 2 sec. At that time

vB/A(t) = vB(2.109) − vA(2.109) = (−54.256) − (54.256) = 108.5 ft/s.

1.134 Solution: vA(0) = 45 mph = 66 ft/s, vB(0) =58 mph = 85 ft/s, xB/A(0) = 500ft, taking xA(0) = 0 then xB(0) = 500 ft, aA = 4 ft/s and aB = −2 ft/s.Integrating each yields

vA(t) = vA(0) + 2t = 66 + 2t, xA(t) = xA(0) + 66t + t2 = 66t + t2,

vB(t) = vB(0) − 2t = 85 − 2t, xB(t) = xB(0) + 85t − t2 = 500 + 85t − t2. Thecars meet when xB/A = 0 or when 500+85t−2t2 = 0. This yields two roots fort, one is negative and the other is t = 33.795s. Using the above formulas, thecars meet at

xA(33.795) = 3, 373 ft ∼ 0.64 miles and their velocities are

vA(33.795) = 133.59 ft/s and

vB(35.795) = 17.41 ft/s

1.135 Given: aB/A = 3 · m/s2, vB/A(0) = 0, xB/A(0) = 10 m, and xA = 3t2 + 1compute aB. Differentiation of xA yields vA = 6t and aA = 6. So that aB/A =

3 = aB − aA = aB − 6, and solving yields aB = 9 m/s2. Integrating aB/A = 3yields vB/A(t) = vB/A(0) + 3t = 3t. Integrating again yields xB/A = vB/A(0) +

3t22 = 10 + 3t2. Since xB/A = xB − xA we have 10 + 3t2

2= xB − 3t2 − 1 or

xB(t) = 4.5t2 + 11 m.

1.136 Since the time is so large in seconds, leave this in mph, miles and hours. Forthe 747, vB = 575 mph and xB = 575t miles. At t = 3 hr., xB(3) = 575(3).At t = 3 the Concorde takes off and vA = 1336 mph so that xA = 1336t miles.For t ≥ 3, xB can be rewritten as xB = (575)(3) + 575t, t ≥ 3. The Concorde

catches the 747 at xA = xB or (575)(3)+575t = 1336t or t = (575)(3)(76)

= 2.27 hrs.

1.137 The pendulum falls as θB(t) = θ0 cos√

gℓt, with θB(0) = 0. Consider the vertical

(θ = 0) position as the reference point at which we would like to know the time

tc, i.e., solving θ0 cos√

gℓt = 0 for t yields

√

gℓtc = π

2or tc = π

2

√

ℓg

= 1.121 s,

the time it takes the pendulum to cover the angular distance from 20◦ to zero(note this does not depend on the value of θ0). Next consider particle A. It

66

starts at rest (xA(0) = 0) 10 meters to the left, so that xA(0) = −10 m. Theparticle’s acceleration can be written as an = a m/s2 where a is an unknownconstant. Integrating yields vA = vA(0) + at = at. Integrating again yields:

xA(t) = xa(0) + a2t2 = −a

2t2 − 10. The value at collision for t will be tc = π

2

√

ℓg

so that a2t2 − 10 = 0 or a = 20

t2c= 80g

π2ℓ= 15.9 m/s2.

1.138 From eq. (1.77) vB = ℓθB = 5θB since the pendulum is in uniform circularmotion or radius ℓ = 5 m. Differentiation of the expression for θB(t) yields

θB = −θ0

√

gℓsin

(√

gℓt)

so that vB = ℓθ − θ0

√gℓ sin

(√

gℓt)

. Let tc be the

time of impact calculated in problem 1.37. Then from 1.37, vA = atc so thatvB/A = −θ0

√gℓ sin

(√

gℓt)

− 20t2c

tc. Substitute tc = π2

√

ℓg

and θ0 = π9

radians (20◦)

yields vB/A = π9

√gℓ + 40

π

√

gℓ

= 20.279 m/s.

1.139 The acceleration of each particle is given by

aA = −9.81 − 0.1vA (1) aB = −9.81 − 0.1vB (2)

with initial conditions vA(0) = 30 m/s, xA(0) = 100 m vB(0) = xB(0) = 0.Equation (1) becomes xA +0.1xA = −9.81 and (2) becomes xB +0.1xB = −9.81

These are second order, uncoupled linear differential equations with analyticalsolutions. The homogeneous solutions are xA = A, a constant and xA = Be0.1t

or xA(t) = A+Be−0.1t. The particular solution is Ct. Substitution of (xA)p = Ctinto (1) yields C = −98.1. Thus (xA)p = −98.1t and the total solution isxA = A + Be−0.1t − 98.1t. Applying the initial condition yields

100 = A + B from the initial position

30 = −0.1B − 98.1 from the initial velocity

Solving yields B = −681 and A = 781. Thus

xA(t) = 781 − 681e−0.1t − 98.1t and vA(t) = 68.1e−0.1t − 98.1.

Next consider particle B which has the same form but different initial conditions,i.e., xB = A′ + B′e−0.1t − 98.1t. The initial conditions yield

0 = A′ + B′

0 = −0.1B′ − 98.1

Solving yields A′ = 981 and B′ = −981 so that

xB(t) = 981(1 − e−0.1t) − 98.1t and vB(t) = 98.1e−0.1t − 98.1

Computing xB/A(t) and setting this equal to zero yields

0 = 981 − 781 + (−981 + 681)e−0.1tc

Solving yields tc = ℓn(2/3)−0.1

= 4.0555, the time of collision. The relative velocityat tc = 4.055 is vB(4.055) − vA(4.055) = 20 m/s

67

The two differential equations for the acceleration of the particles are separableand can be solved by hand. However, to gain a better conceptual understandingof the particle motion, the equations are solved by numerical integration usingthe following Mathcad code:

a v 9.81 .0.1 vi . .0 4060

∆t 0.001

t .i ∆t

vA

0yA

0vB

0yB

0

100

30

0

vAi 1

yAi 1

vBi 1

yBi 1

vAi

.a vAi

∆t

yAi

.vAi

∆t

vBi

.a vB i ∆t

yBi

.vB ∆t

0 1 2 3 4 50

50

100Position- time of A and B

time s

yA

yB

ti

=yA 29.241 =yB 29.252 The impact occurs 29.2 m above theoriginal position of B (taken as the origin).

=vB vA4055

19.999 The relative velocity at impact is 20 m/s

0

i

68

The Matlab code for this is:

function xdot=onept139(t,x);

xdot=[x(2);-9.81-0.1*x(2);x(4);-9.81-0.1*x(4)];

In the command window:

EDU>tspan=[0 4];

EDU>x0=[100;0;0;30];

EDU>[t,x]=ode45(‘onept139’,tspan,x0);

EDU>plot(t,x(:,1),t,x(:,3))

1.140 Let vs denote the velocity of the swimmer starting from rest and vc = 0.1+0.1xs

denote the velocity of the river (opposite of vs). Then vs/c = 1.5 m/s = vs − vc

so that vs = vc + 1.5 = −0.1 − 0.01xs + 1.5 or dxs

dt+ 0.01xs = 1.4 which can

be solved by use of an integrating factor for xs. Here p(t) = 0.01, f(t) = 1.4,

λ(t) = e∫

.01dt = e0.01t so that xs(t) = e0.01t [∫

e0.01t(1.4) + C0] = 140 + C0e−0.01t.

Since xs(0) = 0, C0 = −140 and xs(t) = 140(1 − e−0.01t), xs(60) = 140(1 −e−0.6t) = 63.2 m.

. .0 14000

,x t 1.4 .0.01 x

∆t 0.01

x 0

x

i.v ,x

it ∆t

0 50 100 1500

50

100

150

x

i

ti

i

v

0

xi+1

FIGURE S1.140

69

The Matlab code is:

x(1)=0;dt=0.001,t(1)=0

for n=1:14000

x(n+1)=x(n)+(1.4-0.01*x(n))*dt

t(n+1)=t(n)+dt

end

plot(t,x)

1.141 Since xs(t) = 140(1 − e−0.6t) from problem 1.40, the swimmer will reach 100 m

at time t that satisfies 100 = 140(1 − e−0.6t) or t = −ℓn(1/7)0.1

= 125 s.

1.142 Given vA(t) = j m/s (constant), vB(0) = 0, aB(t) = 2[cos 30◦i + sin 30◦j] n/sa constant. Also rA(0) = 0, rB(0) = −4i + j m. From va(t) = 3j. rA(t) =rA(0) + 3tj = 3tj. Integration of aB(t) yields: vB(t) = vB(0) + 1.73ti + tj m/s= 1.73ti + tj. Integrating again yields

rB(t) = 1.732

t2 i + 12t2j + rB(0) = (0.866t2 − 4)i + (0.5t2 + 1)j

Thus rB/A(t) = (3t + 4 − 0.866t2)i − (0.5t2 + 1)j. For this to be zero, each

component must be zero, which cannot be because the j component is alwayspositive, hence t has no real roots. The distance between the particles is |rB/A| =√

(3t + 4 − 0.866t2)2 + (0.5t2 + 1)2. The minimum can be found from

ddt

[(3 + 4 − 0.866t2)2 + (0.5t2 + 1)2] = 0

2(3t + 4 − 0.866t2)(3 − 1.73t) + 2(0.5t2 + 1)(t) = 0

which is cubic in t and has positive solutions. Then t = 0 or rB/A = 4i = j is

when the two particles are closest. As time evolves they move apart (rootsfound using Mathcad).

1.143 Using y up along north and x to the right along east, the airplane has relativespeed of vA/W = 200(cos θi + sin θj) mph where θ is the unknown direction.

The wind velocity is vw = 50(cos 45◦i + sin 45◦i) = 50√2i + 50√

2j = constant. We

would like vA = 200i. From the definition then

vA = vW + vA/W = (50)[(0.707)i + (0.707j)] + 200(cos θi + sin θ)j

which results in the two scalar equations

vA = 35.4 + 200 cos θ

0 = 35.4 + 200 sin θ

Solving yields: θ = −10.2◦ and vA = 232.2 mph.

70

1.144 Note that the flanker travels 40 ft (20 ft/s × 2 s = 40 ft) before the quarterback throws the ball at t = 0. Let xf(0) = 40i denote the initial position of thequarter back. For the ball

ab = −gk

so that

vb = v cos θ cos β i + (−gt + v sin θ)j + v cos θ sin βk

and

xb = [v cos θ cos βt + xb(0) i + [−gt2

2+ v sin θt + yb(0)]j + [v cos θ sin βt]k

where xb(0) = −30 ft (i.e. 30 yards behind the line of scrimmage) and yb(0) = 6ft. For the flanker xf = 20i, and integrating yields

xf(t) = (20t + xf(0))i + 8j + 60k

Setting xb = xf yields the three equations

from i: 20t + 40 = v cos θ cos βt − 30

from j: 8 = −gt2

2+ v sin θt + 6

from k: 60 = v cos θ sin β.

Solving these three nonlinear equations in the three unknowns t, β, v for θ = 30◦

using a numerical procedure yields

v = 69.719 ft/s, β = 28.152◦ and t = 2.106.

From the expression for xf the yards gained on the play are

(20)(2.106)+403

= 27.375 yd

1.145 Following sample 1.28, let vs be the absolute velocity of the swimmer. With i

perpendicular to the shore and j up stream, vs = vi, where v is the unknownabsolute speed of the swimmer. Let θ be the angle that vs/w (swimmer relativeto the water) so that

vs/w = 6 cos θi + 6 sin θj (ft/s)

vw = −5j ft/s, so that

vs = vw + vs/w becomes

vi = −5j + 6 cos θi + 6 sin θj

from i: v = 6 cos θ

from j: 0 = 6 sin θ − 5

which is 2 equations in two unknowns: θ and v. Solving yields θ = 56.4◦,v = 3.3 ft/s. The time to travel 200 ft is t = d

v= 200

3.3= 60.3 sec.

71

1.146 Solution:

We can form a block model of the United States as shown.

Seattle

21.80

45 0 450

1200 mi

1200 mi 600 mi 1200 mi Miami

The direct flight from Seattle to Miami would be along a path South 21.80 of East. Thewind direction for the 1200 miles was estimated to be South 450 East, during the next 600miles the wind direction is to the East and during the last 1200 miles, the wind is North450 East. The wind speed is estimated to be 50 mph.

The general relative motion equation for any segment of the flight is:v v vP w p w= + /

For the first leg of the flight, these vectors are:

v i j

v i j

v i j

p

w

p w

= −[ ]= −[ ]

= +[ ]

vp cos( . )ˆ sin( . )ˆ

cos( )ˆ sin( )ˆ

cos( )ˆ sin( )ˆ/

21 8 21 8

50 45 45

300 θ θ

The scalar equations of relative motion are solved using the Given-Find function inMathcad:

Make an initial guess for the velocity of the plane and the bearing of the planev p 300

θ .10 degGiven

.v p cos .21.8 deg .50 cos .45 deg .300 cos θ

.v p sin .21.8 deg .50 sin .45 deg .300 sin θ

72

1.147 Solution:

Seattle

21.80

1200 mi

750 mi 250 mi 500 mi 1000 mi 500 mi Miami

Wind Vel E S 85 0 E E N 30 0 E S 300 E

The solution will be made using the Given - Find function of Mathcad:

Leg 1.

Make an initial guess for the velocity of the plane and the bearing of the planev p 300

θ .10 degGiven

.v p cos .21.8 deg .100 cos .0 deg .300 cos θ

.v p sin .21.8 deg .100 sin .0 deg .300 sin θ

=Find ,v p θ390.541

0.505=

0.505

deg28.934

vp = 391 mph θ = -30 0

Leg 2.

Make an initial guess for the velocity of the plane and the bearing of the planev p 300

θ .10 degGiven

.v p cos .21.8 deg .100 cos .85 deg .300 cos θ

.v p sin .21.8 deg .100 sin .85 deg .300 sin θ

=Find ,v p θ331.502

0.078=

0.078

deg4.469

73

1.148 Solution:

Seattle

San Francisco 45 0 450 New York

1200 mi

1200 mi 600 mi 1200 mi Miami

The flight is now to the West against the jet stream. The flight will be divided into threelegs.

Leg 1.

Make an initial guess for the velocity of the plane and the bearing of the planev p 200

θ .10 degGiven

.v p cos .10 deg .50 cos .45 deg .200 cos θ

.v p sin .10 deg .50 sin .45 deg .200 sin θ

=Find ,v p θ167.082

0.032=

0.032

deg1.833

The velocity of the plane and the course bearing are: vp = 167 mph θ = S 2 0 W

Leg 2.

Make an initial guess for the velocity of the plane and the bearing of the planev p 200

θ .10 degGiven

.v p cos .10 deg .50 cos .0 deg .200 cos θ

.v p sin .10 deg .50 sin .0 deg .200 sin θ

=Find ,v p θ150.571

0.131=

0.131

deg7.506

S 10 0 W

74

1.149 Solution:

21.80 Chicago

1200 mi

Memphis

750 mi 250 mi 500 mi 1000 mi 500 mi

Wind Vel E S 85 0 E E N 30 0 E S 300 E

The desired flight path from Chicago to Memphis is assumed to be S 800 E under a 100-mph wind N 300 E.

Make an initial guess for the velocity of the plane and the bearing of the planev p 300

θ .80 degGiven

.v p cos .80 deg .100 cos .30 deg .300 cos θ

.v p sin .80 deg .100 sin .30 deg .300 sin θ

=Find ,v p θ250.701

1.715=

1.715

deg98.262

The velocity of the plane and the course bearing are: vp = 251 mph θ = S 82 0 W

1.150 From sample 1.29 the time to reach the opposite shore is tf = 2dvB/w cos θ

= 40cos θ

sec. The expression for y(t) at y(tf) = 500 becomes

500 = 5[

(

10·cos θ200

)2403

3·cos3 θ− 210 cos θ

200· 402

2 cos2 θ

]

+ 10 sin θ 40cos θ

which is transendental in θ and has solution (see numerical root finder) θ = 56.43◦.

75

1.151 From sample 1.29 the angle is determined

θ = sin−1(

2vw

3vB/w

)

= sin(

(2)(2)(3)(10)

)

= 7.6◦

where vw = 2 ft/s is the velocity of the river.

1.152 From sample 1.30, the 3 equations determining vs, θ and β with vRT = 0.7become

θ + β = 90◦

−vs cos β = −0.3 cos θ

vs sin β = 0.7 − 0.3 sin θ

Solving using Mathcad yields: θ = 24.8◦, β = 64.3◦ and vs = 0.632 m/s.

1.153 We are given: xB = 2 sin πti + 5tj,

xG = 4t cos 30◦i + 4t sin 30◦j so that differentiation yields

vB = 2π cos πti + 5j, vG = 4 cos 30◦i + 4 sin 30j and

aB = −2π2 sin πti, aG = 0

From eq. 1.126, vG/B = vG − vB = (4 cos 30◦ − 2π cos πt)i + (4 sin 30◦ − 5)j

and aG/B = aG − aB = 2π2 sin πti

76

1.154 From the previous problem:

xG/A = (4t cos 30◦ − 2 sin πt)i + (4t sin 30◦ − 5t)j

so that the distance between the two as a function of time is

d(t) = |x(t)| = ((4t cos 30◦ − 2 sin πt)2 + (4t sin 30◦ − 5t)2)1/2

which is plotted below using Mathcad:

θ .30 deg t ..,0 0.1 3

xB t

.2 sin .π t.5 t

0

xG t

..4 t cos θ..4 t sin θ

0d t xB t xG t

0 1 2 3

5

10

15

Time s

Dis

tanc

e m

d t

t

FIGURE S1.154

1.155 Given xB = 2 sin πti + 5tj and

xG = 4t cos 30◦i + 4t sin 30 sin(

πt2

)

j,

xB/G = (2 sin πt − 4t cos 30◦)i + (5t − 4t sin 30◦ sin πt2)j

vB/G = (2π cos πt − 4 cos 30◦)i + (5 − 2πt sin 30◦ cos πt2− 4 sin 30◦ sin πt

2)j

aB/G = −2π2 sin πti + (−4π sin 30◦ cos πt2

+ π2t sin30◦ sin πt2)j

77

1.156 From xB/0 of the previous problem:

d(t) = |xB/G(t)| = [4π4 sin2 πt + (1 − 4π sin 30◦ cos πt2

+ πt sin 30◦ sin πt2)2]1/2

which is plotted in the figure using Mathcad.

θ .30 deg t ..,0 0.1 3

xB t

.2 sin .π t.5 t

0

xG t

..4 t cos θ

...4 t sin θ sin.π t

2

0 d t xB t xG t

0 1 2 3

10

20

30

Time s

Dis

tanc

e m

d t

t

FIGURE S1.156

The Matlab code is:

syms t

th=pi/6;

xb=[2*sin(pi*t);5*t;0];xG=[4*t*cos(th);4*t*sin(th)*sin(pi*t/2);0];

d=sqrt(dot(xb,xG));

ezplot(d,[0,3])

78

1.157 From the given function

aA = ddt

(vA) = i + 1.224j− 1.224k − 0.2vA

which is a vector differential equation of the form

vA + 0.2vA = i + 1.224j− 1.224k = b

The solution of the homogeneous equation vAH= Ae−0.2t plus a particular

solution: vp = 10.2

(i + 1.224j− 1.224k) or

vA = Ae0.2t + 10.2

(i + 1.224j− 1.224k)

where A is evaluated from the listed condition vA(0) = 0 which yields

0 = A + 10.2

(i + 1.224j− 1.224k) or A = − 10.2

(1.224j− 1.224k)

Then

vA = 5(1 − e−0.2t)(i + 1.224j− 1.224k) m/s

Integration again yields

xA − xA(0) = 5(t + 5e−0.2t − 5)(i + 1.224j− 1.224k)

or

xA = 5(t + 5(e−0.2t − 1))(i + 1.224j− 1.224k)

Next consider aB = 3ti − 2 cos πtk m/s. Integrating yields

vB − vB(0) = 3t2

2i − 2 sin πt

πk where vB(0) = 0

Integrating again yields

xB − xB(0) = t3

2i − 2

π2 (1 − cos πt)k where xB(0) = 0

Now form

vB/A =[

3t2

2− 5(1 − e0.2t)

]

i − 6.12(1 − e0.2t)j

+[

6.12(1 − e−0.2t) − 2π

sin πt]

k

and

xB/A =[

t3

2− 5(t + 5(e−0.2t − 1))

]

i − [6.12(t + 5(e−0.2t − 1))] j

+[

1π2 (cos πt − 1) + 6.12(t + 5(e−0.2t − 1))

]

k

79

The figure represents a numerical solution and plots of the magnitude of xB/A

in Mathcad:

i ..0 3000 ∆t 0.001 ti.i ∆t

vx0 0 x0 0 vy0 0 y0 0 vz0 0 z0 0

ax ,vx t 1 .0.2 vx .3 t

ay ,vy t .2 0.0612 .0.2 vy

az ,vz t .0.621 2 .0.2 vz .2 cos .π t

vxi 1

xi 1

vyi 1

yi 1

vzi 1

zi 1

vxi.ax ,vxi ti ∆t

xi.vxi ∆t

vyi.ay ,vyi ti ∆t

yi.vyi ∆t

vzi.az ,vzi ti ∆t

zi.vzi ∆t

vi vxi2 vyi

2 vzi2 xi xi

2 yi2 zi

2

0 1 2 3

5

10

xi

ti

0 1 2 3

5

10

vi

ti

FIGURE S1.157

The Matlab code ode45 can be used to solve this numerically. The plots of x(t)and r(t) can be obtained symbollically from:

sys t

x=[t 3/2-5*(t-5*(exp(-0.2*b)-1));6.12*(t+5*(exp((-0.2*t)-1));(1/pi 2)*(cos(pi*t)-1)];

Nx=SVD(x);

ezplot(Nx(3),[0,3])

80

1.158 Define d(t) = |xB/A| where xB/A is defined as derived in Problem 1.157. UsingMathcad to plot yields figure S1.157.

1.159 Since xA starts to move 1.5s after xB, shift the time of xB ahead by 1.5s byreplacing t with t + 1.5 in the expression for xB derived in Problem 1.157. Thisyields

xB = (t+1.5)3

3i − 1

π2 (1 − cos(π(t + 1.5))k

Then for t ≥ 4.5s,

xB/H =[

(t+1.5)3

2− 5(t + 5(e−0.2t − 1)

]

i − 6.12 [t + 5(e0.2t − 1)] j

+{

6.12 [t + 5(e−0.2t − 1)] − 2π2 (1 − cos(πt + 1.5))

}

k

and for 0 < t < 1.5

xB/A = (t+1.5)3

2i − 2

π2 (1 − cos(π(t + 1.5))k.

The Mathcad solution:

i ..0 3000 ∆t 0.001 ti.i ∆t

vx0 0 x0 0 vy0 0 y0 0 vz0 0 z0 0

ax ,vx t .1 .0.2 vx Φ t 1.5 3 t

ay ,vy t ..2 0.0612 .0.2 vy Φ t 1.5

az ,vz t ..0.621 2 .0.2 vz Φ t 1.5 .2 cos .π t

vxi 1

xi 1

vyi 1

yi 1

vzi 1

zi 1

vxi.ax ,vxi ti ∆t

xi.vx i ∆t

vyi.ay ,vyi ti ∆t

yi.vyi ∆t

vzi.az ,vzi ti ∆t

zi.vzi ∆t

vi vxi2

vyi2

vzi2

xi xi2

yi2

zi2

0 1 2 3

10

20

xi

ti

0 1 2 3

20

40

vi

ti

FIGURE S1.159

81

1.160 This is similar to sample problem 1.32. Using the coordinate system suggestedin Sample 1.32, yB denotes the distance to the bracket and xA to the man. Then

xA + 2yB = ℓ

2vB = −vA or vA = −2vB

Since vB = 1 ft/s, vA = −2 ft/s (down)

1.161 Use the pulley as a reference point, and let xA be the distance from the pulleyto A and xB the distance from the pulley to B. Then

xA + xB = constant and differentiating yields

xA = −vB, and aA = −aB.

1.162 Let vc be the cable velocity: vc = 0.5 m/s. Let xT denote the distance from thetree to the end of the truck. The length to the cable is xc + 2xT = ℓ so that

vc = −2vT or vT = −12vc = −0.25 m/s (right)

1.163 Let xA be the distance for the pulley to the man at A, xp be the distancebetween the top pulley and the bottom pulley. Let xB be the distance from thetop pulley to B and let the bracket for the bottom pulley be a height of c. Then

xA + 2xp + xB − c, but xp = (xB − c)

so that xA + 3xB =constant and vA = −3vB

with xB, xA pointed down, B traveling upwards yields

vB = −10 ft/s and vA = −3 · (−10) = 30 ft/s (down)

1.164 Let xA denote the distance from the top of the pulley to block A, yB denote thedistance from the top of the pulley down to the pulley holding B and y denotethe distance from the top of the pulley down to mass c (down position, to theright position). Then the length of the rope is

yc + 2yB + xA = constant

differentiation yields

vA + 2vB + vC = 0

or

vc = −2vB − 3 m/s and ac = −2aA − 1 m/s

82

1.165 Let h be the hypothesis of the triangle. Then from the right triangle

h2 = 152 + x2 (1)

for any value of x. Also from length of rope

ℓ = h + (15 − y) = 30 (2)

Solve this forh = y + 15 (3)

Now substitute (3) into (1) to get:

(15 + a)2 = 152 + x2 (4)

for any x and y. Taking a time derivative yields

2(15 + y)y = 2xx

Solving for y yields

y =xx

15 + y

At y = 10(15)2 = 152 + x2 or, x = 20 m

Then y =(

2025

) (

0.5 m/s2)

= 0.4 m/s, at y = 10 m.

1.166 Let xA extend from the fixed pulley above B to the left of pulley B to mass Aand let yB extend (positive) from the same pulley down to the mass at B. Letd be the distance between the two fixed pulleys. Then the length of the rope is

ℓ = 2yB + xA + (xA − d) + (xA − d)

or

const. = 2yB + 3xA.

Differentiate to get:

yB = −32

xA and yB = −32xA.

1.167 Use the top fixed pulleys as a reference and let xA denote the distance to massA, xB the distance to mass B, xc the distance to mass C and denote xp thedistance to tohe pulley that is free to move. All distances are vertical withpositive downward. There are two separate ropes. Let ℓ1 denote the from A tothe movable pulley so that

ℓ1 = xA + xB + (xB − xp) (1)

83

Let ℓ2 denote the length of the rope from mass C up to ground so that

ℓ2 = xc + 2xp (2)

Differentiation of (2) yields vp = −vc/2 which is substituted into the derivativeof (1) to yield

4vB + vc = −2vA m/s

1.168 Let the top pulley be the fixed frame of reference, and x denote the distancefrom the reference down (t) to the mass, and let xm be the distance from thereference down (t) to a point on the rope connecting to the motor. Them

ℓ = xm + 2x

and differentiation yields that

v = 12vm

From section 1.7 vm = rw = (0.2)(9.55 rad/s = 1.91 m/s and v = 12vm =

0.955 m/s

1.169 Let yB denote the distance from the pulley to the mass at A (+ down), let xB

denote the distance from the pulley to the collar along the shaft. The length ofthe rope is ℓ = da + yB. From the right triangle made by “d”, the bar the collar

rides on, and the rope: ℓ21. Thus ℓ = d + yB = yB +

√

d2 + x2B. Differentiating

yields 0 = yB + 12

2xB xB√d2+x2

B

.

1.170 Pick a fixed point to left of block A as the reference and define xA and xc

(both positive to the right) as the distance from the reference to block A and Crespectively. Then define yc to be the distance down from the first pulley to B.The length of the rope ℓ is then ℓ = xA + xc + 2yB anad differentiation yieldsvA = −vc − 2vB.

1.171 Solution:dsdt

= 20 m/s = constant so s = 20t and ds2

dt2= 0

θ(s) = 4 sin( s2000

) so that θ(t) = 4 sin(0.5t)

From equation 1.147 with x0 = y0 = 0

x(s) =∫ s0 cos(θ(η))dη y(s) =

∫ s0 sin θ(η)dη

84

The Mathcad solution:

s ..,0 10 6000θ s .4 sin

s

2000

x s d0

sucos θ u

y s d0

susin θ u

an s .202 dd s

θ s

4000 3000 2000 1000 0 1000

1000

1000

2000Road as viewed from above

km

km

y s

x s

0 2000 4000 6000

0.5

1

distance along road in km

norm

al a

ccel

erat

ion

m/s

^2

an s

s

FIGURE S1.171

From (1.146): an =(

dsdt

)2 dθ(s)ds

= (20)2 4200

cos(

s200

)

= 800 cos(

s200

)

y(s) and |an(s)| are plotted in the figure.

1.172 We are given that a(s) = 5 so that integrating vdv = 5ds yields v(s) =√

10sfor zero initial condition. Also since θ = 1 + s2

20002 ,dθds

= 2s10002 . But an(s) =

v(s)2 dθds

= (10s)( 2s20002 ). Thus |a| =

√

a2n + a2

t =√

(10s)2( 2s20002 )2 + 25. When

|a| = 8 the car spins out so that s must satisfy

82 = (10s)2( 2s20002 )2 + 25.

Solving for s yields: s2 =√

39(2000)2

20or s = 1, 118 m.

85

1.173 The given constant speed means that dsdt

= 60 mhour

· hour3600s

5280 ftm

= 88 ft/s. Todetermine the normal acceleration from equation (1.146) we need to calculate

(ds/dt)2 = 882 and dθ/ds. From the given value of θ: dds

(

s2000

+ e−5/1000)

=1

2000− 1

1000e−s/1000 Thus an = 882

1000

[

0.5 − e−s/1000]

= 7.744[

12− e−s/1000

]

ft/s2.

The Mathcad solution:v 88 s ..,0 10 5000

θ ss

2000e

s

1000an s .v

2 d

d sθ s

0 1000 2000 3000 4000 5000

2

4Magnitude of Normal Acceleration

distance traveled in f

ft/s

^2 an s

s

FIGURE S1.173

1.174 This requires the use of the 3-D formulation of equation 1.150. The componentof a normal to the above can be found by taking the dot product of the accel-eration with the unit normal vector. To define the unit normal vector can befound from an = a = at = a − a · t. First we need dθ

dsand dβ

ds. From the given

form: dθds

= 11000

(0.5 − e−s/1000), dβds

= π2

5000cos

(

πs5000

)

. The Mathcad solution:

v 88 s ..,0 10 5000

β s .π sin.π s

5000θ ss

2000e

s

1000

an s .v2 .d

d sθ s

2

cos β s2 d

d sβ s

2

0 1000 2000 3000 4000 5000

10

20

distance traveled in f

no

rma

l a

cc

ele

rati

on

in

ft/

s^2

an s

s

FIGURE S1.174

86

1.175 Solution:

v 88 s ..,0 10 5000 β s .π sin.π s

5000θ ss

2000e

s

1000

v s ..2 1.5 s

an s .v s 2 .dd s

θ s2

cos β s 2 dd s

β s2

0 1000 2000 3000 4000 5000

20

40

distance traveled in ft

norm

al a

ccel

erat

ion

in ft

/s^2

an s

s

FIGURE S1.175

1.176 Solution:

v 88 s ..,0 10 5000 β s .π sin.π s

5000θ ss

2000e

s

1000

ρ s1

.dd s

θ s2

cos β s 2 dd s

β s2

0 1000 2000 3000 4000 5000

1000

2000

3000

distance traveled in ft

radi

us o

f cu

rvat

ure

in f

t

ρ s

s

FIGURE S1.176

87

1.177 Solution:

s ..,0 1 50 θ ss

100 β s .π2

sin.π s

25

v 10

an s .v2 .dd s

θ s2

cos β s 2 dd s

β s2

0 10 20 30 40 50

10

20

distance traveled in m

norm

al a

ccel

erat

ion

in m

/s^2

an s

s

FIGURE S1.177

1.178 Solution:

s ..,0 1 100 θ ss

100 β s .π2

sin.π s

25

v s 10 .5 cos.π s

25

an s .v s 2 .dd s

θ s2

cos β s 2 dd s

β s2

at s .dd s

v s v s

a s at s 2 an s 2

0 20 40 60 80 100

20

40

60

distance traveled in m

tota

l acc

eler

atio

n in

m/s

^2

a s

s

FIGURE S1.178

88

CHAPTER 2

2.1 Given: µk = 0.1, W = 200 lb and P = 50 lb. The free body diagram yields:

N − mg − P sin 30 = 0 (sum of forces in y-direction) (1)

P cos 30 − f = max (sum of forces in x-direction) (2)

From (1) N = mg + P sin 30 = 200 + 50(12) = 225 lb

Friction force f = µkN = 0.1(225) = 22.5 lb

Therefore max = 50(√

32

) − 22.5 = 20.80 lb

⇒ ax = 20.80200/32.2

= 3.34 ft/s2

⇒ vx = 3.34t (assuming initially at rest)

⇒ x = 1.67t2 (assuming x(0) = 0)

Therefore t =√

1001.67

= 7.73 sec

F_

30o

f

N

x

y

FIGURE S2.1

2.2 Solution: µs = 0.4, m = 150 kg, µk = 0.3

T − f − mg sin θ = max

N − mg cos θ = may = 0

⇒ N = mg cos θ = 150(9.81) cos θ = 150(9.81) 100√1002+12

= 1471 N

T

mg100

f

N

x

o (1/100) = .573 0.01 rad~~θ = tan-1

θ

1

FIGURE S2.2

90

fmax = µsN = 0.4(1471) = 588.4 N

Therefore Tmax = fmax + mg sin θ = 588.4 + (150)(9.81) 1√1002+1

= 603 N

f = µkN = 0.3(1471) = 441 N

ax = 1m

(T − f − mg sin θ) = 1150

[

603 − 441 − 150(9.81) 1√1002+1

]

ax = 0.98 m/s2

Therefore vx = 0.98t m/s assuming zero initial velocity.

Therefore t = 50.98

= 5.1 sec

2.3 Examining the maximum friction µs = 0.2. This can produce an acceleration ofac = 0.2(32.2) = 6.44 ft/s2. Therefore the box slips because the force of staticfriction mac is less than that applied by the truck

ac = 0.15(32.2) = 4.83 ft/s2

ac/T = 4.83 − 10 = −5.17 ft/s2

(We could now determine where the crate falls off if we knew length of tuckbed.)

mg

f

N

FIGURE S2.3

2.4 Solution:

a)∑

F = ma

100 − 50 = 10032.2

a

a = 16.1 ft/s2

b) 100 − T = 10032.2

a

T − 50 = 5032.2

a

Therefore 50 = 15032.2

a ⇒ a = 10.73 ft/s2

c) 150 − T = 15032.2

a

91

T − 100 = 10032.2

a

⇒ 50 = 25032.2

a ⇒ a = 6.44 ft/s2

100 10050 50

150100

100

50 T

100

T

50

T

150

T

100

FIGURE S2.4

2.5 Solution:

N − mg = ma, or

N = mg + ma

for: a = +2 : N↑ = m(9.81 + 2) = 1181 N

for: a = −2.5 : N↓ = 100(9.81 − 2.5) = 731 N

N

+

mg

FIGURE S2.5

2.6 Solution:

−mg sin 30◦ + µmg cos 30◦ = ma along the conveyor. Or:

−0.5g + 0.2598g = a, thus:

a = −.24g.

92

v = −0.24gt + v0

x = −0.12gt2 + 0.2t

0.12(9.81)t2 − 0.2t − 10 = 0

t = 3s

Note that this slips immediately as µsg cos(30◦) < 0.5g

30o

10 9.8

30o

.

N

f

FIGURE S2.6

2.7 From S2.7a the constraint equation is

2(xA − xB) + (xc − xB) = constant

Thus 2xA − 3xB = constant and differentiation yields

2aA = 3aB (1)

Now from S2.7b, Newton’s law in the horizontal direction yields

3T = (40)aB (2)

and from S2.7c, Newton’s law yields

100 − 2T = (20)aA (3)

This is 3 equations in 3 unknowns which yields

aB = 1.765 m/s2, aA = 2.647 m/s2 and T = 23.533 N

C

X B

B A

x

B A3T

40g

N

100

20g

2T

N

(a) (b) (c)x

A

FIGURE S2.7

93

2.8 Constraints to motion are from figure S2.8a are :

xc + xp = ℓ1, therefore: ap = −ac.

And xA − xp + xB − xp = ℓ2, hence:

aA + aB − 2ap = 0, and thus aA + aB + 2ac = 0

C

P

B

A

x

FIGURE S2.8a

From the FBD’s of figure S2.8b:

25g − T2 = 25aA

T1 = 2T2

20g − T2 = 20aB

50g − 2T2 = 50ac.

But ac = −12(aA + aB)

Subtract 2x the first equation from third to get

50(ac − aA) = 0, or ac = aA.

From the constraint aB = −3aA. Subtract 2nd from 1st

5g = 85aA → aA = 0.577 m/s2

T2 = 231 N aB = −1.731 m/s2

T1 = 462 N ac = 0.577 m/s2

T2 T2 T1 T1

25g 20g2T2

50g

FIGURE S2.8b

94

2.9 From figure S2.9a, the constraints are

xA + 2xp = ℓ1, and

(xB − xp) + xB = ℓ2.

A 100 50

P

B

xA

xP

xP

FIGURE S2.9a

Differentiating yields aA = −2ap and 2aB = ap so that aA = −4aB

Next the FBD’s of Figure S2.9b yields:

100 · 9.81 − T1 = 100aA from A

T2 − 2T1 = 0 (assumed massless) from P

(50)(9.81) − 2T2 = 50aB from B

T1 T1 T12T

2

100(9.81) T2 50(9.81)

T2 =2T1

A P B

FIGURE S2.9b

These three equations, plus the constraints aA = −4aB form a system of fourequations and four unknowns which can be solved for aA. This yields

aA = 8.324 m/s2

Integrating vdv = aAdx from rest conditions yieldsv2

2= (9.324)(0.5) or v = 2.885 m/s

2.10 From S2.10, (100)a = 100x or

x = −50x

vdv = −50dxv2

2= −25(x2 − 0.32)

v =√

50(0.09 − x2)

dxdt

= 7.07√

0.09 − x2

95

∫ t0.3

dx√0.09−x2

=∫ t0 7.07dt

sin−1(

x0.3

)

|x0.3 = 7.07t

sin−1(

x0.3

)

− sin−1(1) = 7.07t

sin−1(

x0.3

)

= (7.07t + π2)

x = 0.3 sin(7.07t + π2)

x = 0.3 cos(7.07t)

Fs

x

FIGURE S2.10

Alternately from S2.10

−kx = mx where x(0) = 0, x(0) = 0.3m

m = 2 kg, k = 100 N/m, x(0) = 0.3

x + 50x = 0

Integrating Factor

x = Aeat, a2 + 50 = 0, a = ±7.07i

x = A sin(7.07t) + B cos(7.07t)

x(0) = 0.3 = B → B = 0.3

x(0) = 0 = 7.07A → A = 0

x = 0.3 cos(7.07t)

2.11 From the FBD of the system as a whole: P − 3mg sin α = 3ma, so that

a = 13m

[P − 3mg sin α]

From the FBD for car C:

P − TBC − mg sin α = ma = 13[P − 3mg sin α].

Solving yields

TBC = 2/3P .

Likewise a FBD of the system consisting of B and C taken together yields

TAB = 13P

96

A

B

C

mg

α

α

mg

mg

TBC

P

C

mg

P

FIGURE S2.11

2.12 From a FBD of the entire system as given in the top of Figure S2.12:

70 sin 20◦ − 70 cos 20◦(0.2) = 7032.2

a

a = 4.96 ft/s2

Now from the bottom of S2.12, the FBD yields

50 sin 20◦ − T − 0.2(50) cos 20◦ = 5032.2

(4.96)

So T = 0!

This result is expected, as the motion is independent of mass and both blocksslides independently.

20

20

o

20

50

o

20o

f

T50

FIGURE S2.12

2.13 Sled will start to move when

a) 500(t2 + 2t) = 0.5(200)(9.81) t2 + 2t − 1.962 = 0

t = −1 ±√

4+4(1.962)

2= −1 ±

√2.962

t = 0.721 s or t = −2.721

97

F(t)

f

FIGURE S2.13

b) From t = 0.721 to 6s

F (t) − 0.4(200)(9.81) = 200a

v = 1200

∫ 60.721[500(t2 + 2t) − 0.4(200)9.81]dt

= [0.833t3 + 2.5t2 − 3.924t]|60.721

= 248 m/s (555 mi/hr)

v(t) = 0.833t3 + 2.5t2 − 3.924t + 1.217

x(6) = [0.208t4 + 0.833t8 − 1.962t2 + 1.217t|60.721

x(6) = 386 m

c) From 6s on:

−0.4g = a

a = −3.924

v = −3.924t + 248

time where v = 0

t = 63.2s

x = −3.9242

t2 + 248t

= 7837 m

Total distance = 386 + 7837 = 8223 m or 8.2 km (5.1 mi)

98

2.14 Solution: fmax = 0.3(90)(9.81) = 265 N. Therefore system slips.

a) Assume whole system slips as a whole. Then summing forces yields: 1200−(.25)(990)(9.91) = (90)(a). Solving yields a = 10.88 m/s2.

20

40

30

90(9.81)

f

1200N

A + B + C

220.7

1200

a=10.9 m/s2

60f=60a a=2.9 m/s

2A+B

f=0.3(60)(9.81) = 176.58

A+B slips

FIGURE S2.14a

b) Assume slip between each surface

20g

f 2

N 1

N1

f2

N 2

f1

30g

f2 N2

1200

f3

40g

N 3

FIGURE S2.14b

0.25(20)(9.81) = 20aA from FBD of the 20kg block. Thus

aA = 2.45 m/s2

0.25(60)(9.81)− 0.25(20)9.81) = 40aB from the FBD of the 40kg block thus

aB = 2.45 m/s2

Therefore A and B the top two blocks move together (they have the sameacceleration)

99

c) Next C slips out from under A and B (the FBD is S2.14c). Then

0.25(60)(9.81) = 60aAB. Thus

aAB = 2.45 m/s2.

From the FBD on the bottom of Figure S2.14c:

1200 − 0.25(60)(9.81)

−0.25(90)(9.81) = 30ac

Thus the acceleration of the bottom block is

ac = 27.74 m/s2

60g

fAB

NAB

NAB

fAB

1200

f

N

30g

C

C

FIGURE S2.14c

2.15 Will the system slip?

20

40

30 400

FIGURE S2.15a

Is 400 ≥ 0.3(90)(9.81) = 265 N?

Yes, so the system slips.

A

B

f max=176.6

a=2.9 m/s2N

mg

FIGURE S2.15b

100

From the FBD of A and B:

400 − 0.25(90)9.81 = 90aT

aT = 1.99 m/s2

fmax = 176.6 which implies a = 2.9 m/s2

So A and B must move with C.

2.16 Solution:

N1

f1

30o

N2

f2

50g

N2

f2 f3

N3

10g

FIGURE S2.16a

Assume no slip and determine the required value of µs. From FBD of A:

N1 + µN2 cos 30◦ − N2 sin 30◦ = 0 (1)

µN1 + µN2 sin 30◦ + N2 cos 30◦ − 50g = 0 (2)

From the FBD of B:

−µN2 cos 30◦ + N2 sin 30◦ − µN3 = 0 (3)

−µN2 sin 30◦ − N2 cos 30◦ + N3 − 10g = 0 (4)

This is a system of four nonlinear equations in the four unknowns µs, N1, N2,and N3. Solved by Mathcad Eqs. 1-4 required µs = 0.242.

Since the µ required for equilibrium is larger than the 0.2 given the systemwill slip. Knowing it will slip, the equations of motion are (with zero accelera-tion into the walls)

N1 + µkN2 cos 30◦ − N2 sin 30◦ = 0

µkN1 + µkN2 sin 30◦ + N2 cos 30◦ − 50g = 50aA

−µkN2 cos 30◦ + N2 sin 30◦ − µkN3 = 10aB

−µkN2 sin 30◦ − N2 cos 30◦ + N3 − 10g = 0

µk = 0.15, this is a system of 5 unknowns N1, N2, N3, aA, aB and only 4equations. Hence we need a constraint of the motion. The acceleration of the

101

blocks in the normal direction at the contacting surface is the same (see S2.16b).Thus −aA cos 30◦ = aB sin 30◦ which provides the 5th equation. Solving these 5nonlinear equations in the 5 unknowns yields the desired acceleration (using acomputer code such as the MATLAB file at the end of this solution):

aA = −3.401 m/s2

aB = 5.891 m/s2

From d = a2t2, the time for box A to move 0.1 m is

t =√

(2)(0.1)3.401

= 0.242 s. During this time box B moves

xB(0.242) = 5.8912

(0.242)2 = 0.173 m

θθ

aA

aB

-aA cos 30o = aB

sin 30o

FIGURE S2.16b

This must be added to the distance the block B slides to the right due to itsseparation velocity. The velocity of separation is:

vB = 5.891 · 0.242 = 1.426 m/s

The deceleration of the block is −0.15 · 9.81 = −1.472 and the time when theblock stops is:

t = 1.4261.472

= 0.969 s

The total distance traveled is:

xB = −0.075 · 9.81(0.969)2 + 1.426 · 0.969 + 0.173 = 0.864 m

102

The MATLAB solution:

Main program to solve the first nonlinear equation(‘non1.m’)

% Main m-file to solve the nonlinear equation

% Initial guess values for the solution

% x(1)=N1; x(2)=N2; x(3)=N3; x(4)=Mu_s;

g=9.81;

x0=[ 10*g 6*g 10*g 0.4];

% solve the nonlinear equation with given initial guess

F=fsolve('neqn1',x0);

% display solutions

disp(' ')

disp(sprintf(' N1= %6.4f',F(1)))

disp(sprintf(' N2= %6.4f',F(2)))

disp(sprintf(' N3= %6.4f',F(3)))

disp(sprintf(' Mu_s= %6.4f',F(4)))Function program to solve the first nonlinear equation(‘neqn1.m’)

function F=neqn1(x)

% Function m-file to define the nonlinear equation

deg2rad=pi./180; g=9.81; % constants required for the calculation

N1=x(1); N2=x(2); N3=x(3); Mu_s=x(4); % assign the values

F1= N1 + Mu_s * N2 * cos(30*deg2rad) - N2 * sin(30*deg2rad) ;

F2= Mu_s * N1 + Mu_s * N2 * sin(30*deg2rad) + N2 * cos(30*deg2rad) - 50*g ;

F3= -Mu_s * N2 * cos(30*deg2rad) + N2 * sin(30*deg2rad) - N3 * Mu_s ;

F4= -Mu_s * N2 * sin(30*deg2rad) - N2 * cos(30*deg2rad) + N3 - 10*g ;

F=[F1 F2 F3 F4]';Main program to solve the second nonlinear equation(‘non2.m’)

% Main m-file to solve the nonlinear equation

% Initial guess values for the solution

% x(1)=a_A; x(2)=a_B; x(3)=N1; x(4)=N2; x(5)=N3;

g=9.81;

x0=[ -0.3 0.4 10*g 6*g 10*g ];

% solve the nonlinear equation with given initial guess

F=fsolve('neqn2',x0);

% display solutions

disp(' ')

disp(sprintf(' a_A= %6.4f',F(1)))

disp(sprintf(' a_B= %6.4f',F(2)))

disp(sprintf(' N1= %6.4f',F(3)))

disp(sprintf(' N2= %6.4f',F(4)))

disp(sprintf(' N3= %6.4f',F(5)))Function program to solve the first nonlinear equation(‘neqn2.m’)

function F=neqn2(x)

% Function m-file to define the nonlinear equation

deg2rad=pi./180; g=9.81; Mu_k=0.15; % constants required for the calculation

a_A=x(1); a_B=x(2); N1=x(3); N2=x(4); N3=x(5); % assign the values

F1= N1 + Mu_k * N2 * cos(30*deg2rad) - N2 * sin(30*deg2rad) ;

F2= Mu_k * N1 + Mu_k * N2 * sin(30*deg2rad) + N2 * cos(30*deg2rad) - 50*g - 50 * a_A;

F3= -Mu_k * N2 * cos(30*deg2rad) + N2 * sin(30*deg2rad) - N3 * Mu_k - 10 * a_B;

F4= -Mu_k * N2 * sin(30*deg2rad) - N2 * cos(30*deg2rad) + N3 - 10*g ;

F5= a_A * cos(30*deg2rad) + a_B*sin(30*deg2rad);

F=[F1 F2 F3 F4 F5]';

103

2.17 The geometry of the sketched spring in Fig. S2.17 yields

ℓ2 = ℓ20 + x2

(ℓ0 + ∆)2 = ℓ20 + x2

ℓ20 + 2ℓ0∆ + ∆2 = ℓ2

0 + x2

where ∆ is the spring elongation.

m

x

l

l

o

Fs

N

mg

FIGURE S2.17

Thus

∆2 + 2ℓ0∆ − x2 = 0, or

∆ =−2ℓ0±

√4ℓ2

0+4x2

2= −ℓ0 ±

√

ℓ20 + x2. Using the positive root, the magnitude

of the spring force is

|Fs| = k[

√

ℓ20 + x2 − ℓ0

]

From the FBD:

mg − k[

√

ℓ20 + x2 − ℓ0

]

x√ℓ20+x2

= ma

Thus

d2xdt2

= g − km

[

√

ℓ20 + x2 − ℓ0

]

x√ℓ20+x2

a(x) = g − kxm

(

1 − ℓ0√ℓ20+x2

)

.

2.18 Solution:

From the figure

(ℓ + ∆)2 = x2 + ℓ2

ℓ2 + 2ℓ∆ + ∆2 = x2 + ℓ2

∆2 + 2ℓ∆ − x2 = 0

∆ = −2ℓ±√

4ℓ2+4x2

2=

√ℓ2 + x2 − ℓ (could write this directly)

mg − Fs sin θ − µkN = mx, sin θ = x√ℓ2+x2

N − Fs cos θ = 0, cos θ = ℓ√ℓ2+x2

104

θ

θ

Ν

F s

f

mg

x

l

FIGURE S2.18

N = k[√

ℓ2 + x2 − ℓ]

ℓ√ℓ2+x2

, Fs = k∆

N = kℓ[

1 − ℓ√ℓ2+x2

]

x = g − km

[

x(

1 − ℓ√ℓ2+x2

)

+ µkx|x|ℓ

(

1 − ℓ√ℓ2+x2

)]

x = g − km

(x + µkx|x|ℓ)(1 − ℓ√

ℓ2+x2)

which would require a numerical solution.

105

2.19 Solution:

l 0.3 k 300 m 2 g 9.81

i . .0 2000

∆t 0.001

v0

x0

0

0

a x g ..k

ml2 x2 l

x

l2 x2

vi 1

xi 1

vi

.a xi

∆t

xi

.vi

∆t

0 0.5 1 1.5 20.2

0

0.2

0.4

0.6Position vs time

Time s

xi

.i ∆t

2.20 First note that 30g ≥ 0.5(20)g, so it does in fact slip. The constraint is that

L = xA + xB + const

⇒ 0 = xA + xθ

⇒ 0 = aA + aB so that aA = −aB . (1)

From the FBD of (B):

30(9.8) − T = 30aB (2)

From the FBD of A:∑

Fx = maA becomes:

T − µkN = 20aA

T − µkmg = 20aA

T − 0.4(20)(9.81) = (20)aA (3)

Adding (2) and (3), using (1) yields

(30)(9.81) − (0.4)(20)(9.81) = 50aA

106

So that aA = 4.316 m/s2. Then d = aB

2t2 or

t =√

14.326

= 0.481 s.

20g

+

T

N

F

A

T

B

30g

+

(a) (b)

FIGURE S2.20

2.21 First determine the unit normal vector to the surface of the ice, n.

100 ft

z

x

y

45

30

t T

n

FIGURE S2.21

Then compute the component of the weight vector along n, denoted N . Theforce parallel to the ice will then be S = W −N , where W is the weight. Oncethe unit normal along S is determined, the motion can be treated as rectilinearalong the face of the ice. The following Mathcad code completes the solution.

107

T

cos .45 deg

0

sin .45 deg

t

0

cos .30 deg

sin .30 deg

W

0

0

130

nT t

T t

=n

0.655

0.378

0.655

Let us find the component of the weight that acts normal to the face of the ice.

N ..W n n

=N

55.714

32.167

55.714

The force parallel to the ice face causing her to slip is S

S W N

=S

55.714

32.167

74.286

We can first see if she slips by comparing |S| to µk|N|

=S 98.271 =.1.0 N 85.105

Therefore she slips.

The unit vector in the direction of the slip is s

sS

S

We can now treat the motion as rectilinear along the ice face in the direction s.

a .32.2

130S .0.8 N

=a 7.477 ft/s2

The distances she slides in the s direction is:

d = −100(t · s) = 66.144 ft

Now her velocity after sliding 66.144 ft can be determined.

v dvdx

= a, v =√

2ad, v = 31.45 ft/s. Yes, she will be injured.

108

2.22 Constraint: xA + xB = ℓ therefore aA = −aB

x B

x A

20

60

30o

Constraint

xA+xB

l=

... aA = -a B

FIGURE S2.22

From the FBD of A:

20g sin 30 + 0.2(20g) cos 30◦ − T = 20aA

From the FBD of B:

60g sin 30◦ − 0.2(20g) cos 30◦ − 0.1(80g) cos 30◦ − T = 60aB

Thus

132.1 − T = 20aA

192.4 − T = 60aB

Which along with the constraint equation is a system of 3 equations in 3 un-knowns. Solving yields

aB = 0.753 m/s2

aA = −0.753 m/s2

aA/B = aA − aB = −1.506 m/s2

A T

fA

NA

m Ag

(a)

B

m Ag

N B

f B

T

fA

N A

(b)

FIGURE S2.22a,b

109

2.23 The rope length yields: 2xA + yB = ℓ, so that 2aA + aB = 0

150

8 kgx

y

A 3 kg

2xA+ y

B= l

2aA+ aB = 0

FIGURE S2.23

A FBD of A yields:

150 − 2T = 8aA (1)

A FBD of B yields:

3 · (9.81) − T = 3aB (2)

Using the constraint, the 1st equation becomes

150 − 2T = −4aB

Multiply (2) by 2 yields

58.86 − 2T = 6aB

Subtracting these last two equations yields

−10aB = 91.14, so aB = −9.114 m/s, and aA = 4.557 m/s2

Also from (2): T = 3(9.81) − 3aB = 56.8 N

Solving

vB = aBt

for t yields:

t = 2.54.557

= 0.549 s

2.24 The constraint here is that xA + xB = constant, so that aA = −aB . Assumethat the system moves to the left.

From FBD of mass B

for y direction: NB − mBg cos β = 0, so NB = mBg cos β and fB = µkNB =µkmBg cos β

from x direction: −T + mBg sin β − fB = mBaB

Thus we have: aB = g(sinβ − µk cos β) − TmB

(1)

110

aB

+

T

f B = N B

N Bm

Bg

(a)

mA

g

aA+

T

fA= NA

NA

(b)

µk

µk

αβ

FIGURE S2.24a,b

From the FBD of mass A

from the y-direction: NA − mAg cos α = 0 or NA = mAg cos α

from the x-direction: −T + mAg sin α − µkmAg cos α = mAaA

Thus aA = xA = g(sin α − µk cos α) − TmA

. (2)

Now use the constraint aA+aB = 0 along with equations (1) and (2) to eliminatethe tension T . This yields

g(sinα − µk cos α) − TmA

+ g(sin β − µk cos β) − TmB

= 0

Solving yields T = g(

mAmB

mA+mB

)

(sin α + sin β − µk(cos α + cos β))

Thus the value of constant acceleration from (2) is:

xA = g(sinα − µk cos α) − g(

mAmB

mA+mB

)

[sin α + sin β − µk(cos α + cos β)]

2.25 Solution:

m

mga

y

x

N

µ s N

θ

FIGURE S2.25

From the FBD, the sum of forces in the x direction yields:

N − mg = ma sin θ

111

The force sin in the y direction is:

µsN = ma cos θ

Thus

µs(mg + ma sin θ) = ma cos θ

or

µsg = a(cos θ − µs sin θ)

or

a = µsgcos θ−µs sin θ

.

2.26 Solution:

m

mga

N

θ

f

x

y

FIGURE S2.26

From the FBD, the force sum in the x direction is:

µsN = ma cos θ

The force sum in the y direction is:

mg − N = ma sin θ

Thus:

N = m(g − a sin θ)

and

µsm(g − a sin θ) = ma cos θ

µsg = a(cos θ + µs sin θ)

a = µsgcos θ+µs sin θ

112

2.27 From the top drawing, the constraints are

B

A

x

y

2T1

20g

T1 2T2

3T 2

FIGURE S2.27

2yA + xp = ℓ1 and xB + 2(xB − xp) = ℓ2. Thus 3aB + 4aA = 0 (1)

From the FBD of the pulley, T1 = 2T2. Thus the FBD of A yields:

20(9.81) − 4T2 = 20aA (2)

The FBD of B yields:

−3T2 = 10aB → −3T2 = −1043aA so T2 = 40

9aA (3)

Substitution of (3) into (2) yields:

20(9.81) − 4(

409

)

aA = 20aA

or

(180 + 160)aA = 9(20)9.81

and

aA = 5.194 m/s2.

xA(t) = aA

2t2 thus

1.5 = 5.1942

t2, T2 = 23.1 N, t = 0.76 s, T1 = 46.2 N

2.28 From the drawing, the constraints are:

xA + xB + (xB − xp) = ℓ2 and 2xp + xc = ℓ1 combining yields

xA + 2xB + xc

2= const

Thus: 2aA + 4aB + ac = 0. Let T1 be the tension in cable 1 and T2 = T1/2 bethe tension in cable 2.

113

A

B

C

x

T1

10g

2T1

80g

2T 2T2

20gT1

A B CP

FIGURE S2.28

Now since a = 12at2, t =

√

2da

=√

(2)(4)4.5

= 1.33 s.

From the FBD of A: 10g − T1 = 10aA

From the FBD of B: 80g − 2T1 = 80aB

From the FBD of C: 20g − T1

2= 20ac

Rearranging these 4 equations in 4 unknowns yields:

10aA + T1 = 10g,

80aB + 2T1 = 80g

20ac + 12T1 = 20g

2aA + 4aB + ac = 0

In matrix form these become:

aA

aB

ac

T1

=

10 0 0 10 80 0 20 0 20 0.52 4 1 0

−1

10 · 9.8180 · 9.8120 · 9.81

0

which can be solved using a code to yield:

aA = −11.319, aB = 4.528, aC = 4.528,

T1 = 211.292 and T2 = T1

2= 105.646

Now since a = 12at2, t =

√

2da

=√

(2)(4)4.5

= 1.33 s.

2.29 Solution:

Does system slip? 100 > 30(9.81)0.2 = 58.9 so it slips.

From the FBD of the system as a whole:

100 − 0.15(30)(9.81) = 30aT

Thus

114

N

f

20

10 100

FIGURE S2.29

aT = 1.862 m/s2

From the FBD of B 0.2(20)9.81 = 20aB or aB = 1.962 which is larger than aT

so there is enough friction for the system to move as one.

2.30 Solution:

20g

0.15 N B

NB

0.15N B NB

10g 100

0.15 NA

NA

FIGURE S2.30

From B

(0.15)(20)g = 20aB or aB = 1.47 m/s2

From A

100 − (0.15)(20)g − (0.15)(30)g = 10aA

So

aA = 2.643 m/s2

Block B slips.

2.31 Solution:

Note that aA = a cos θi + a sin θj and aB = xB i + yB j

so that

115

B

A

a

30o

NB

30g

y

x

NA30o

50g

NB

FIGURE S2.31

aB/A = aB − aA = (xB − a cos θ)i + (yB − a sin θ)y

Since aB/A cannot have a component in the j direction we have the constraintthat

yB = a sin θ (1)

From the FBD of B we have

(x direction) 0 = mBxB so that xB = 0

(y direction) NB − mBg = mB yB (2)

From the FBD of A we have

(x direction) −NA sin θ = mAxA = mAa cos θ (3)

(y direction) −NB − mAg + NA cos θ = mAa sin θ (4)

This yields a system of 4 equations in the 4 unknowns yB, a, NA and NB whichcan be written in matrix form as

1 − sin θ 0 030 0 0 −10 50 cos θ sin θ 00 50 sin θ − cos θ 1

yB

aNA

NB

=

0−(30)(9.81)

0−(50)(9.81)

This has solution

a = −6.824 m/s2, yB = −3.411 m/s2, NA = 591.0 N, NB = 191.9 N.

The relative acceleration is

aB/A = (xB − a cos θ)i

which has magnitude

0 + (6.824)(.866) = 5.91 m/s2.

Thus

t =√

2da

=√

2(.1)5.91

= 0.184 s

116

2.32 Solution:

y

x

mg

N

Fs

α

l

α

FIGURE S2.32

The spring force is

Fs = k∆.

From the geometry

∆ =√

ℓ2 + x2 − ℓ.

Summing forces in the x direction (see S2.32) yields

mg cos α − k(√

ℓ2 + x2 − ℓ) x√ℓ2+x2

− µkNx|x| = mx

Summing forces in the y direction yields:

N − mg sin α − k(√

ℓ2 + x2 − ℓ) ℓ√ℓ2+x2

= 0

Therefore

x = 1m

{

mg cos α − kx(

1 − ℓ√ℓ2∗x2

)

− µkx|x|

[

mg sin α + kℓ(

1 − ℓ√ℓ2+x2

)]}

or

x = 1m

{

mg cos α −(

kx + kℓµkx|x|

) (

1 − ℓ√ℓ2+x2

)

− µkx|x|mg sin α

}

117

2.33 Solution:

k 40:= µk 0.2:= L 0.2:= m 5:= g 9.81:= α 30 deg⋅:=

i 0 4000..:=

∆t 0.001:=

v0

x0

0.00001

0

:=

a x v,( )1

mm g⋅ cos α( )⋅ k x µk L⋅

v

v⋅+

⋅ 1L

L2

x2

+

−

⋅− µk m⋅ g⋅ sin α( )⋅v

v⋅−

⋅:=

vi 1+

xi 1+

vi

a xi

vi

,( )∆t⋅+

xi

vi∆t⋅+

:=

0 1 2 3 40

1

2

3

xi

i ∆t⋅

118

2.34 Solution:

x

50 ft

(a)

50x

x

110

(b)

FIGURE S2.34a,b

First determine her velocity when the bungee cord becomes taut

x = +g, but vdv = +gdx, so that v0 =√

2 · g · ℓ = 62.16 ft/s

Now the FBD is illustrated in S2.34b and yields −kx + mg = mx, with v = v0.

Thus a = v dvdx

= g − km

x

Integrating both sides yields:∫ 0v0

vdv =∫ d0

[

g − km

x]

dx

−gℓ = gd − k2m

d2

d =−g±

√g2+ 2kgℓ

m−km

= 77.893 ft (positive root)

Therefore 150 − 60 − 77.9 = 12.1 ft above the ground.

Also, note from the quadratic formula above (using k = c/ℓ) that d is propor-tional to ℓ. Thus, Fx = kd = cd

ℓis independent of ℓ. For the numbers given,

Fx = 5 · 77.9 = 389.5 lb.

2.35 Solution:

cv

mg

y

x

FIGURE S2.35

From the FBD, the sum of the forces in the y-direction when the skydiver hasreached her terminal velocity (i.e. zero acceleration) yields:

cv − mg = 0

119

Solving for c = 4.0A = mgv

.

Thus A = 14.0

· (60)(9.81)9

= 16.4 m2

2.36 From the FBD of the man there are 3 forces acting on him: gravity, the dragforce, which shuts “off” after 10 meters, and the buoyancy force which shuts offwhen he stops. The force summation yields

−Fb − Fc + mg = mx

Now Fc = cv from 10 m or Fc = cv < x − 10 >0

Likewise FB = −mg from 13m or < x − 13 >0

Thus

x = − cvm

< x − 10 >0 −g < x − 13 >0 +g

Fc Fb

mg+

FIGURE S2.36

The force at impact is mx| = mg − cv, where v =√

2ad and x = 10, thus

Fimpact = 70 · 9.81 − 500√

(2)(10)(9.81) = −6313 N. The depth beneath the

surface that the diver travels is d = 13.195-10 =3.195 m (see Mathcad soln).

m 70:= c 500:= ∆t 0.001:= g 9.81:= i 0 5000..:=

a v x,( ) gc v⋅

mΦ x 10−( )⋅

v

v⋅− g Φ x 13−( )⋅−:=

v0

x0

0

0

:=v

i 1+

xi 1+

vi

a vi

xi

,( )∆t⋅+

xi

vi∆t⋅+

:=

0 5 10 150

10

20

vi

xi

x3057

13.194=

x3058

13.195=

x5000

13.195=

120

2.37 This can be solved by “hand” or by using the singularity function and numeri-cally integrating.

By hand: for the first 5 seconds:

a(t) = 0.7t,

v(t) = 0.72

t2

s(t) = 0.76

t3

At t = 5, we have v5 = 8.75 m/s, s5 = 14.58 m.

After the first 5 seconds a = −2 m/s and integrating yields (starting the clockover at t = 0)

v(t) = −2t + v5 = −2t + 8.75

s(t) = −2t2

2+ 8.75t

Solving v(t) = 0 for t yields t = 8.752

= 4.38s.

Thus in the second interval s(4.38) = 19.14 m is reached.

The total distance traveled is 14.58 + 19.14 = 33.72 m.

This is solved by simple numerical integration in the following Mathcad code.

The total distance traveled is 33.8 meters.

i . .0 940

∆t 0.01

g 9.81 m 1200

t .i ∆t

a t ..Φ 5 t 0.7 t .Φ t 5 2

v

0x

0

0

vi 1

xi 1

vi

.a ti

∆t

xi

.v ∆t

0 5 10

20

20

40

vi

xi

ti

0

i

i

121

2.38 The free body diagram given in S2.38:

+

Fk

Fd

mg

FIGURE S2.38

Here Fd = cx < x − 35 >0< v > is the damping force applied by the matand Fk = k(x − 35)2 < x − 35 >0 is the spring force applied by the mat. Theequation of motion obtained by summing forces in the vertical direction is justmx = mg − Fd − Fk. Two design criteria must be met: total force must beless than 600 lb and the total mat displacement must be less than 6 ft. Thefollowing Mathcad code computes the displacement by numerical integration,and can be evaluated for various values of c and k simply by changing thedefinition statements for these constants. The solution is shown for one possiblecombination of c and k that satisfy the design criteria.

g 32.2:= m120

g:= k 32:= c 20:=

a v x,( ) g Φ x 20−( )k

m⋅ x 20−( )

2⋅− Φ x 20−( )

c

m⋅ v⋅ Φ v( )⋅−:=

∆t 0.001:= i 0 8000..:= ti

i ∆t⋅:=

v0

x0

0

0

:=v

i 1+

xi 1+

vi

a vi

xi

,( )∆t⋅+

xi

vi∆t⋅+

:=

Fi

m a vi

xi

,( )⋅:=

0 5

500

0

Fi

i ∆t⋅

0 5 100

20

xi

ti

122

2.39 The sum of forces in the vertical direction ignoring gravity is

mx = 50000[< 3 − t >0 +∑4

n=1 < t − 6n >< 6n + 3 − t >]

The following Mathcad code integrates the equation of motion and display thedisplacement and velocity:

m 500:= F 50000:=

a t( )F

mΦ 3 t−( )

1

4

n

Φ t 6 n⋅−( ) Φ 6 n⋅ 3+ t−( )⋅( )∑=

+

⋅:=

∆t 0.01:= i 0 2700..:= ti

i ∆t⋅:=

v0

x0

100

0

:=v

i 1+

xi 1+

vi

a ti( )∆t⋅+

xi

vi∆t⋅+

:= v2700

1.604 103

×=

x2700

2.3 104

×=

0 10 20 30

0

2000

vi

ti

0 10 20 300

2 .104

xi

ti

2.40 A FBD of the ball is given in S2.40a.

20o

y

x

mg

F

FIGURE S2.40

The problem is solved by setting up the computational solution and then per-forming a trial and error procedure to find the required force. The Mathcadsolution is shown for one possible value of F (roughly the minimum possiblevalue). The initial conditions are vx(0) = −120 ft/s, x(0) = 0, y(0) = 4 ft (thisis assumed to be the level of the ball when it is struck, based on an averagebatter size and general location of the strike zone) and vy(0) = 0. The forcerequired is found to be about 128 lb.

123

W5

16:= g 32.2:= m

W

g:=

F 13200 m⋅:= F 128.106=

ax t( )F

mΦ 0.015 t−( )⋅ cos 20 deg⋅( )⋅:= ay t( ) g−

F

mΦ 0.015 t−( )⋅ sin 20 deg⋅( )⋅+:=

∆t 0.001:= i 0 4550..:=

ti

i ∆t⋅:=

vx0

x0

vy0

y0

120−

0

0

4

:=

vxi 1+

xi 1+

vyi 1+

yi 1+

vxi

ax ti( )∆t⋅+

xi

vxi∆t⋅+

vyi

ay ti( )∆t⋅+

yi

vyi∆t⋅+

:=x4534

354.065=

y4534

7.137 104−

×=

0 100 200 300 4000

50

100

y i

xi

2.41 A free body diagram of the box is given in the figure.

The forces are Fk = kx < x − 4 >0, f = µk cos θ v|v|

The FBD yields the following two equations∑

Fy : N − mg cos 45◦ = 0∑

Fx : mx = −f − Fk + mg sin 45◦

124

f

N

mg

y

Fk

FIGURE S2.41

The following Mathcad code integrates to find the solution.

µ:= 0.6

i . .0 5000 ∆t 0.001

m 2 g 9.81 k 500θ .45 deg

t .i ∆t

a ,v x .1

m..m g sin θ ..µ cos θ

v

v.Φ x 4 .k x 4

0 1 2 3 4 5

5

5

vi

xi

t

v

0x

0

0

0v

i 1

xi 1

v

i

.a ,v

i

x i ∆t

x .v ∆t

ii

i

i

125

The equivalent MATLAB code is:

function yprime = ds2pt41(t,y) yprime(1,:) = y(2,:);

% friction forceif y(2) = = 0,

Ef = 0;else

Ef = 4.162*y(2,:)/abs(y(2,:)):end

% spring forceif y(1)<=0,

Fk = 0;else Fk = 250*y(1,:);end

yprime(2,:) = -EF-Fk+6.947;

return

2.42 Working with the free body diagram of the previous problem replace the forcesFk with Fk + Fc where Fc = cv(x − 4). Thus the equation of motion can bewritten as: ma = mg(sin θ − µk cos θ v

|v|)− < x − 4 >◦ (k(x − 4) + cv)

The following MATLAB code can be used to solve for the velocity and displace-ment using ODE for plotting the response.

function yprime = ds2pt41(t,y)

yprime(1,:) = y(2,:);

%friction forceif y(2) = = 0,Ff = 0;elseEf = 4.162*y(2,:)/abs(y(2,:));end

% spring and damper forceif y(1) < =0,Fc = 0;Fk = 0;elseFc = 12*y(2,:);Fk = 250*y(1,:);end

yprime (2,:) = -Ef-Fc-Fk+6.947;

return

126

Next the Mathcad solution is presented, complete with the plots:

k 500:= µk 0.6:= m 2:= g 9.81:= θ 45 deg⋅:=

i 0 5000..:= ∆t 0.001:= ti

i ∆t⋅:=

a v x, c,( )1

mm g sin θ( ) µk cos θ( )⋅

v

v⋅−

⋅

⋅ Φ x 4−( ) k x 4−( )⋅ c v⋅+[ ]⋅−

⋅:=

v0

x0

0

0

:=v

i 1+

xi 1+

vi

a vi

xi

, 20,( )∆t⋅+

xi

vi∆t⋅+

:=

0 2 4 60

2

4

6

xi

ti

v0

x0

0

0

:=v

i 1+

xi 1+

vi

a vi

xi

, 200,( )∆t⋅+

xi

vi∆t⋅+

:=

0 2 4 60

2

4

6

xi

ti

These codes are run several times for different values of c until one is foundthat keeps the box from losing contact with the spring. A value of c = 20 verynearly keeps it in contact, but a value as high as c = 200 is needed to truly keepthe box in contact with the spring.

127

2.43 Assuming positive in the direction of the expanding bag:

30 m/s

vB

FIGURE S2.43

Fd = 5 Ns/m ·vH/B m/s

vH/B = 30 + vB

5(30 + vB) = 500

vB = 70 m/s

2.44 Solution:

Fa Fs

mg

FIGURE S2.44

From the free body diagram, Newton’s law in the vertical direction yields F =mg − 10H(x− 5)(x − 5) − 0.5|v|v Where the Heaviside function indicates thatthe spring forces does not act until x = 5.

The MATLAB code for solving this consists of the following m-file, then usingODE and PLOT to compute and plot the solution x(t). The m-file is

function yprime = ds2pt41(t,y)

yprime(1,:) = y(2,:);

% viscous forceFd = 0.1*y(2,:)*abs(y(2,:));

% spring forceif y(1) < =0,Fk = 0;elseFk = 2*y(1,:);end

yprime (2,:) = -Fd-Fk+9.81;

return

128

The Mathcad equivalent is given next along with the plot of x(s) vs. t:

m 5 k 10 c 0.5 g 9.81

i . .0 10000

∆t 0.001

t .i ∆t

a ,v x .m g ..k Φ x 5 x 5 ..c v v

v

0x

0

0

vi 1

xi 1

v

i

.a ,v

i

xi ∆t

x .v ∆t

0 2 4 6 8 100

5

10

15Position vs time

time s

xi

.i ∆t

ii

i

0

2.45 Solution:

ρ 130

N

FIGURE S2.45

ρ = 100 ft v = 60 mph = 88 ft/s∑

Fn = man = mv2

ρ

N − 130 = 13032.2

v2

100

129

Therefore N − 130 =(

13032.2

)

882

100

N = 443 lb up

2.46 Solution:

mg

500 ft

FIGURE S2.46

To fly the hill N = 0

mg = m v2

500, v =

√32.2 × 500 = 126.9 ft/s = 86.5 mph

2.47 Solution:

mg

ρf

Nβ β

FIGURE S2.47

N cos β − mg − f sin β = 0, N sin β + f cos β = mv2

p

f = µsN N = mgcos β−µ sinβ

mv2

ρ= mg (sinβ+µ cos β)

(cos β−µ sin β)

v =√

ρg sinβ+µ cos βcos β−µ sinβ

2.48 Solution: 25 mph = 36.7 ft/s. Therefore 36.72

100= 32.2

[

sinβ+0.3 cos βcos β−0.3 sinβ

]

0.418(cosβ − 0.3 sinβ) = sin β + 0.3 cos β

β = 6.02◦

130

2.49 Solution:

v

Pmg

L

FIGURE S2.49

amax = 5g = v2

pv = 700 mph = 1027 ft/s

ρ = (1027)2

5(32.2)= 6551 ft

2.50 Solution:

β β

mg

FIGURE S2.50

mg sin β = mRβ wdw = gR

sin βdβ

w2

a|ww0

= gR(− cos θ)β

0

w2 = 2gR

(1 − cos β) + w20

At separation:

mg cos β = mv2

Rv2

R2 = 2gR

(1 − cos β) +v2

0

R2

g cos β = 2g(1 − cos β) +v2

0

R3g cos β = 2g +

v2

0

R

β = cos−1[

23

+v2

0

3gR

]

131

2.51 Solution:

N

mg

θ

θ

FIGURE S2.51

The following equations can be written:

mg cos θ = mRθ (1)

N − mg sin θ = mRθ2 (2)

Eq (1) can be rewritten as ω dωdθ

= gR

cos θ (where ω = θ).

Integrating this gives:ω2

2= g

R(sin θ − sin θ0)

Thus:

ω =√

2gR

(sin θ − sin θ0)

and

v =√

2gR (sin θ − sin θ0)

2.52 Solution:

mRθ = mg cos θ − f θ|θ|

−mg sin θ + N = mRθ2

Therefore mRθ = mg cos θ − µkθ|θ| [mg sin θ + mRθ2]

Canceling the mass

θ = gR

cos θ − µk

Rθ|θ| [g sin θ + Rθ2]

132

2.53 Solution:

The mass cancels out of the equation of motion and values have been assumed for theradius and the coefficient of kinetic friction. The Mathcad code is:

i . .0 2000

∆t 0.001 g 9.81 µ k 0.15 R 0.3

t .i ∆t

α ,ω θ .g

Rcos θ ..

µ k

R

ω

ω.g sin θ .R ω

2

ω0

θ

0

.10 deg

ωi 1

θi 1

ωi

.α ,ωi

θ ∆t

θ .ωi

∆t

0 0.5 1 1.5 20

1

2

3Angular Position vs time

Time s

θi

ti

i

0

i

i

The equivalent code in MATLAB is:

function yprime = ds2pt53(t,y)

g = 9.81; R = 1; gg = g/R; mu = 0.07;

yprime(1) = y(2);

if y(2)==0yprime(2) = -mu*(y(2) 2+gg*sin(y(1)))+gg*cos(y(1));elseyprime(2) = -mu*(y(2) 2+gg*sin(y(1)))*y(2)/abs(y(2))+gg*cos(y(1));end

return

133

2.54 See the free body diagram given in Sample Problem 2.12. It is given that:

Fs = kR(2 cos θ2− 1)

Ff = µk|N | θ|θ|

From Sample 2.12, the FBD yields upon summing forces in the normal andtangential directions∑

Fn = −N + mg sin θ + Fs cos θ2

= mRθ2

N = mg sin θ + kR(2 cos θ2− 1) cos θ

2− mRθ2

∑

Ft = −Ff + Fs sin θ2− mg cos θ = mRθ

The equation of motion is thus:

mRθ = −µk

∣

∣

∣mg sin θ + kR(2 cos θ2− 1) cos θ

2− mRθ2

∣

∣

∣

θ|θ|

+kR(2 cos θ2− 1) sin θ

2− mg cos θ

θ = −µk

∣

∣

∣

gR

sin θ + km

(2 cos θ2− 1) cos θ

2− θ2

∣

∣

∣

θ|θ|

+ km

(2 cos θ2− 1) sin θ

2− g

Rcos θ

When the slider is in equilibrium, θ = 0 and θ = 0. The Mathcad solutionfollows.

The solution shown here is for a coefficient of kinetic friction of 0.18, a valuethat was chosen for a typical response. The spring would be unloaded whenθ = 120◦ but there would still be the gravitational acceleration acting on theslider. It may be observed that the slider is oscillating about a slighter higherangle (roughly 123◦), which is the equilibrium angle.

Note the slow start of the slider as it overcomes the friction force. The springforce is high when θ = 30◦ but this contributes to a high normal force andtherefore a high friction force. If the coefficient of kinetic friction is too high,could the problem be such that friction drives the motion? No, this is animpossibility. In this formulation the friction term in the equation of motion ismultiplied by θ/|θ| to make sure the friction always opposes motion; this term iszero when the angular velocity is zero so friction will never initiate motion. Todetermine whether motion would occur in a particular case, determine whetherthe spring force minus the gravitational force in the initial configuration isgreater than the coefficient of static friction times the normal force. That wouldbe a separate calculation as the friction force in this equation of motion is zerountil motion starts.

134

m 1:= k 600:= R 0.2:= g 9.81:= µk 0.18:=

N ω θ,( ) m g⋅ sin θ( )⋅ k R⋅ 2 cosθ

2

⋅ 1−

⋅ cosθ

2

⋅+ m R⋅ ω2

⋅−:=

α ω θ,( ) g−

Rcos θ( )⋅

k

m2 cos

θ

2

⋅ 1−

⋅ sinθ

2

⋅+µk

m R⋅N ω θ,( )⋅

ω

ω⋅−:=

i 0 15000..:= ∆t 0.0001:= ti

i ∆t⋅:=

ω0

θ0

0

π

6

:=ωi 1+

θ i 1+

ωi α ωi θ i,( )∆t⋅+

θ i ωi ∆t⋅+

:=

0 0.5 1 1.50

100

200

θi

deg

ti

f θ( ) g−

Rcos θ( )⋅

k

m2 cos

θ

2

⋅ 1−

⋅ sinθ

2

⋅+:=

x 120 deg⋅:=root f x( ) x,( )

deg123.35=

135

2.55 We are given α = constant and so ω = αt.

r 2ω

rα

FIGURE S2.55

f = m(rαet + rω2en) is the total force. The total acceleration is

µsmg = m√

(rα)2 + (rω2)2

µs = 1g

√

(rα)2 + (r(αt)2)2 = rαg

√1 + α2t4

2.56 Solution:

A

ω_

B

RA

RB

FIGURE S2.56

RA + RB = ℓ

T = mBRBω2 , T = mARAω2

Therefore mBRB = mARA.

RA = ℓ − RB

RB = mA

mBℓ − mA

mBRB

RB = mA

mA+mBℓ

RA = mB

mA+mBℓ

136

2.57 Solution:

R

e

θ

<

t

e

<

n

e<

n

mg

N

FIGURE S2.57

Sum forces along t: N cos θ−mg = 0 sum forces along n : N sin θ = mRω2 sin θ

Therefore N = mRω2, mRω2 cos θ = mg, θ = cos−1(

gRω2

)

.

2.58 First-determine the required velocity at top.

mg

N

n

<

FIGURE S2.58a

From the figure S2.58a:

N + mg = mv2

R, N = 1

2mg

v =√

32gR or ωT =

√

32

gR.

θ

θ

R

ω0mg

N

FIGURE S2.58b

137

From S2.58b:

mRθ = −mg sin θ (sum of tangential forces)∫ ωTω0

ωdω =∫ π0

gR

sin θdθ

ω2

T−ω2

0

2= + g

Rcos θ|π0 = −2g

R

ω20 = 4g

R+ 3

2gR

= 112

gR

v0 =√

112gR

2.59 From Problem 2.58

n

<

N

mg

FIGURE S2.59

N − mg =mv2

0

R= m · 11

2g or N = mg + 11

2mg

N = 132

mg

6.5 times body weight could realistically cause spinal injuries.

138

2.60 See free body diagram shown in the solution to problem 2.61 (S2.61).

From the solution to problem 2.61:

yA = ℓθ2 sin θ+g cos θ sin θγ+sin2 θ

,

where

γ = mA/mB

For ℓ = 0.2 m, θ = 30◦, θ(0) = 0, mA = 3 kg and mB = 2 kg:

yA = 2.427 m/s (acceleration of block A).

The acceleration of block B is:

aB = aAj + ℓθ cos θj− ℓθ sin θi

From the solution to problem 2.61:

(mA + mB sin2 θ)θ + mB θ2 cos θ sin θ + (mA + mB)gℓsin θ = 0

For the numbers given:

θ = −35.036 rad/s2.

And:

aB = 2.427j− 6.068j + 3.504i = 3.504i− 3.641j m/s.

139

2.61 The FBD’s are given in the figure:

m Ag

y

mB

g

(t)θ

θ

TN

m A gT

mB

g

θx

FIGURE S2.61

In the y-direction for block A:

(1) T sin θ = mAyA

and in the radial and tangential directions for block B:

(2) −T + mBg cos θ = mB(−ℓθ2 + yA sin θ)

(3) −mBg sin θ = mBℓθ + mB yA cos θ

Substitution of (2) into (1) yields

sin θ{mBℓθ2 − mByA sin θ + mBg cos θ} = mAyA

or

ℓθ2 sin θ + g cos θ sin θ = (mA

mB+ sin2 θ)yA

Thus we have

(4) yA = ℓθ2 sin θ+g cos θ sin θγ+sin2 θ

.

where

γ = mA/mB

Substitute 4 into 3:

ℓθ + ℓθ2 sin θ+g cos θ sin θγ+sin2 θ

· cos θ = −g sin θ.

(γ + sin2 θ)θ + θ2 cos θ sin θ + gℓcos2 θ sin θ + γ g

ℓsin θ + g

ℓsin3 θ = 0

which can also be written as:

(1 + sin2 θγ

)θ + 1γθ2 cos θ sin θ + g

γℓcos2 θ sin θ + g

ℓsin θ + g

γℓsin3 θ = 0

When mA >> mB (γ very large) this becomes the pendulum equation:

θ + gℓsin θ = 0

Resubstituting for γ into the original equation and simplifying:

(mA + mB sin2 θ)θ + mB θ2 cos θ sin θ + (mA + mB)gℓsin θ = 0

140

2.62 The free body diagrams are given in S2.62:

mA

g

N

T

β

T

mBg

y

x

A: B:

FIGURE S2.62

From the solution to problem 2.63 for θ, with θ = 0 and θ = 0:

θ = −g sinβ cos β(mA+mB)ℓ[mA+mB sin2(β]

Writing the equation of motion in the x-direction for block A in the coordinatesystem shown in S2.62:

mAaA = mAg sin β + T sin β

Writing the equation of motion in the y-direction for block B in the coordinatesystem shown in S2.62:

T cos β − mBg cos β = mBℓθ sin β

Solving the previous equation for T :

T = mBg + mBℓθ tanβ

Substituting this gives the acceleration of block A (which is in the x-direction):

aA = mA+mB

mAg sin β + mB

mAℓθ sin β tan β

And the acceleration of B is:

aB = aAi + ℓθ cos β i + ℓθ sin β j

141

2.63 Solution:

mA

g

NT

β

θ

θ

T

mBg

y

x

A: B:

FIGURE S2.63

T sin(β + θ) + mAg sin β = mAaA

−T sin θ = mBaBx

T cos θ − mBg = MBaBy

aB = aA + aB/A

aA = aA(cos β i − sin β j) + ℓθ2(− sin θi + cos θj)

(1) T sin(β + θ) + mAg sin β = mAaA

(2) −T sin θ = mB[aA cos β + ℓθ cos θ − ℓθ2 sin θ]

(3) T cos θ − mBgg = MB[−aA sin β + ℓθ sin θ + ℓθ2 cos θ]

Therefore aA = TmA

sin(β + θ) + g sin β

Substituting into (2) and (3)

2) −T sin θ = mB

[

TmA

sin(β + θ) cos β + g sin β cos β + ℓθ cos θ − ℓθ2 sin θ]

T[

sin θ + mB

mAsin(β + θ) cos β

]

= −mBgsinβ cos β − mBℓθ cos θ + mBℓθ2 sin θ

T = −mB [g sin β cos β+ℓθ cos θ−ℓθ2 sin θ]

sin θ+mBmA

sin(β+θ) cos β

3) T cos θ = mBg − mB[aA sin β − ℓθ sin θ − ℓθ cos θ]

T [cos θ + mB

mAsin(β + θ) sin β] = mB[q(1 − sin2 β) + ℓθ sin θ + ℓθ2 cos θ]

T = mB [g(1−sin2 β)+ℓθ sin θ+ℓθ2 cos θ]

cos θ+mBmA

sin(β+θ) sinβ

Equating two expressions for T

−[cos θ + mB

mAsin(β + θ) sin β] · [g sin β cos β + ℓθ cos θ − ℓθ2 sin θ]

= [sin θ + mb

mAsin(β + θ) cos β] · [g cos2 β + ℓθ sin θ + ℓθ2 cos θ]

Therefore g sin β cos β cos θ + ℓθ cos2 θ − ℓθ2 sin θ cos θ

+g cos2 β sin θ + ℓθ sin2 θ + ℓθ sin θ cos θ

+mB

mAsin(β + θ)[g sin2 β cos β + ℓθ cos θ sin β − ℓθ2 sin θ sin β]

+mB

mAsin(β + θ)[g cos2 β cos β + ℓθ sin θ cos β + ℓθ2 cos θ cos β]

142

g sin(β + θ) cos β + ℓθ + mB

mAsin(β + θ)[g cos β + ℓθ sin(β + θ)+ ℓθ2 cos(β + θ)] = 0

g sin(β + θ) cos β(mA+mB

mA) + ℓθ[1 + mB

mAsin2(β + θ)]

+ℓθ2 mB

mAsin(β + θ) cos(β + θ) = 0

θ = −g sin(β+θ) cos β(mA+mB)+ℓθ2mB sin(β+θ) cos(β+θ)ℓ[mA+mB sin2(β+θ)]

2.64 Solution: (in Mathcad)

i . .0 2000

∆t 0.001

t .i ∆t

m A 3

m B 2

g 9.81

β 0L 0.5

ddθ ,θ dθ

...m A m B g sin β θ cos β ....m B L dθ sin β θ cos β θ

.L m A.m B sin β θ

2

dθ

0θ

0

.20 deg

dθi 1

θi 1

dθi.ddθ ,θ

idθ ∆t

θ .dθ ∆t

0 0.5 1 1.5 240

20

0

20

40Angle- time relationship

time s

θi

deg

ti

i

i

i

i

0

2

143

2.65 Solution: (in Mathcad)

i . .0 2000

∆t 0.001

t .i ∆t

m A 3

m B 2

g 9.81

β .30 degL 0.5

ddθ ,θ dθ

...m A m B g sin β θ cos β ....m B L dθ sin β θ cos β θ

.L m A.m B sin β θ

2

dθ

0θ

0

.20 deg

dθi 1

θi 1

dθ

i

.ddθ ,θi

dθ

i

∆t

θ .dθ ∆t

0 0.5 1 1.5 2100

50

0

50Angle- time relationship

time s

θi

deg

ti

i

0

ii

2

2.66 Solution:

an = 3 · g = v2

pv = 200 mph = 293 ft/s

p = (293)2

96.6= 889 ft

From the solution of 2.47:

an = g(sin β+µ cos β)cos β−µ sinβ

.

3(cos β − µ sin β) − (sin β + µ cos β) = 0

−(3µ + 1) sin β + (3 − µ) cosβ = 0

Therefore tanβ = 3−µ3µ+1

µ = 0.85

β = 31.2◦.

144

2.67 The Mathcad code to solve this problem is:

N 917:= n 0 N..:= ∆t 0.001:= tn

n ∆t⋅:= L 3:= g 9.81:=

v0

s0

θ0

dθ0

0

0

70− deg⋅

0

:=

vn 1+

sn 1+

θn 1+

dθn 1+

vn

g sin θn( )⋅ 0.3 g cos θn( )⋅ dθn vn( )

2⋅+

⋅+

∆t⋅−

sn

vn∆t⋅+

70− deg⋅ 1

sn

L

3

−

⋅

3 70⋅ deg⋅s

n( )2

L3

⋅

:=

sN

3.001= vN

4.227= tN

0.917=

N 817:= n 0 N..:= ∆t 0.001:= tn

n ∆t⋅:= L 3:= g 9.81:=

v0

s0

θ0

dθ0

0

0

70− deg⋅

0

:=

vn 1+

sn 1+

θn 1+

dθn 1+

vn

g sin θn( )⋅ 0.0 g cos θn( )⋅ dθn vn( )

2⋅+

⋅+

∆t⋅−

sn

vn∆t⋅+

70− deg⋅ 1

sn

L

3

−

⋅

3 70⋅ deg⋅s

n( )2

L3

⋅

:=

sN

2.999= vN

6.668= tN

0.817=

The child’s velocity is 4.227 m/s at the bottom of the slide when the coefficientof kinetic friction is 0.3, and 6.668 m/s when there is no friction. It takes thechild 0.917 seconds to reach the bottom of the slide when the coefficient ofkinetic friction is 0.3, and 0.817 seconds when there is no friction.

145

The MATLAB code to solve the problem is as follows:

function yprime = ds2pt67(t,y)

s = y(1); %positions = y(2); %speed

L = 3; %length of slidemu = 0.3; %kinetic coefficient of friction

g = 9.81; %acceleration due to gravity

theta = -1.222*(1-(s/L) 3); %angletheta = 1.222*3 *(s/L) 2/L; %dtheta/ds

% differential equationsyprime(1,:) = s ;yprime(2,:) = - mu*(s 2*theta + g*cos(theta))*sign(s )-g*sin(theta);

% v x and v yyprime(3,:) = s *cos(theta);yprime(4,:) = s *sin(theta);

return

146

2.68 The Mathcad code to solve the problem:

The height of the slide must first be determined numerically as shown below.

x 0 0

The height was originally choosen to be 3 m, and was reduced to 2.854 m so that thebottom would be 0.6 m from the ground.

y 0 2.854

s . .,0 0.1 3

θ s ..70 deg 1s

3

3

x s x 0 d0

s

ζcos θ ζ

y s y 0 d0

s

ζsin θ ζ

=y 3 0.6

0 0.5 1 1.5 20.5

1

1.5

2

2.5

3Slide Contour

horizontal distance m

y s

x s

The solution for the child falling from the top is straight forward as the only force actingon the child is gravitational acceleration.

y(t) = -g m/s

: 2

y(t) = - gt m/s.

y(t) = + 2.854 m

The time when the child hits the ground is t = 0.763s and the velocity is 7.485 m/s.

vert

ica

l dis

tan

ce m

- gt2

2

147

2.69 Solution:

N 6263

g 9.81

µ 0.5

α .47 degL 46i . .0 N

∆t 0.001

t .i ∆t

a ,,θ dθ v .g sin θ .µ .g cos θ .v dθ

x

0y

0

20

=s6263

46.005

0 10 20 30 4010

0

10

20Slide Profile

Horizontal Position m

yi

xi

The slide provides a very small velocity at the bottom and the time on the slide is 6.263 s.A longer time on the slide may be obtained by changing the initial angle and thecoefficient of friction as well as the length of the slide.

v0

s0

θ

0dθ

0

0

α

0

0

vi 1

si 1

θi 1

dθi 1

vi.a ,,θ dθ

i

v ∆t

si

.v t

.α 1s

i

L

2

..2 αs

i

L2

i i i

∆

=v 3.4536263

=t 6.2636263

0 xi 1

yi 1

x

i

.cos θi si 1

si

y .sin θi

si 1

si

i

i

2

148

2.70 Solution:

θ(s) = (s/25000) cos(s/500), for 0 < s < 3000dsdt

= 100 km/hr = 1003.6

= 27.8 m/s

dθds

= 125000

[cos(s/500) − s500

· sin( s500

)], N(s) = cos(θ(s)) + v2

gdθds

∆s 1:= i 0 3000..:= si

i ∆s⋅:=

θ s( )s

25000cos

s

500

⋅:=

x0

y0

0

0

:=xi 1+

yi 1+

xi

cos θ si( )( )∆s⋅+

yi

sin θ si( )( )∆s⋅+

:=

0 500 1000 1500 2000 2500 3000100

50

0

50

y i

xi

dθ s( )1

25000cos

s

500

⋅s

25000sin

s

500

⋅1

500⋅−:=

v 1001000

3600⋅:= v 27.778=

Ni

cos θ si( )( ) v

2

9.81dθ s

i( )⋅+:=

0 500 1000 1500 2000 2500 3000

1

1.02

Ni

xi

149

2.71 Solution:

s . .,0 10 7000

θ s .s

4000cos

s

4001

dθ s .1

4000cos

s

400.

s

400sin

s

400v 66

g 32.2a max

.0.5 g

a n s .v dθ s

0 1000 2000 3000 4000 5000 6000 700020

0

2018.941213

15.547668

a n s

a max

.7 1030 s

Yes, the occupant slips near the end of the ride.

1

4000

2.72 Solution:

0 < s < 20

θ = s sin−1[0.2 − 0.5 cos( s10

)]

s = 1− m/s∑

Fn = F = ms2θ′

Fmax = µsN = µsmg > ms2θ′

µsg > s2θ′ no slip condition

µs > s2θ′

g= 1020.2571

9.81= 2.62

θ = 5 sin−1[0.2 − 0.5 cos( s10

)]

θ′ = 5 1√1−u2

u′

where u = 0.2 − 0.5 cos( s10

)

u′ = 0.5 sin( s10

) · 110

max θ′(s) = 0.2571 r/m

@ s = 17.17 m

150

The following is a Mathcad analysis:

v 10:= s 0 20..:=

θ s( ) 5 asin 0.2 0.5 coss

10

⋅−

⋅:=

dθ s( )sθ s( )

d

d:=

an s( ) v2

dθ s( )⋅:=

0 10 200

20

40

an s( )

s

µ s( )an s( )

9.81:=

µ 17( ) 2.62=

2.73 Differentiate the given r and θ:

r(t) = 0.300 + 0.100 cos(πt)

r(t) = −0.100π sin(πt)

r(t) = −0.100π2 cos(πt)

θ(t) = π6

sin(πt)

θ(t) = π2

6cos(πt)

θ(t) = −π3

6sin(πt)

v(t) = −0.100π sin(πt)er + (0.300 + 0.100 cosπt)π2

6cos2(πt)eθ

a(t) = [−0.100π2 cos(πt) − (0.3 + 0.1 cos(πt))π4

36cos2 πt]er

+[(0.3 + 0.1 cosπt)(−π3

6sin πt) + 2(−0.1π sin(πt))(π2

6cos πt)]eθ

F = marer + maθ eθ

= m{−0.100π4

36cos3(πt) − 0.300π4

36cos2(πt) − 0.100π2 cos(πt)}er

−0.3mπ3

6sin(πt){1 + cos(πt)}eθ

151

2.74 It is given that:

F(t) = (3t2 − 1)er + cos(πt6)eθ, m = 1

Therefore upon comparison with eq. (2.39) in the book:

m(r − rθ2) = 3t2 − 1

m(rθ + 2rθ) = cos(πt6).

This gives the following differential equation to solve:

r = rθ2 + 1m

(3t2 − 1)

θ = −2 rrθ + 1

mrcos(πt

6)

With initial conditions

r(0) = 2 θ(0) = 0 r(0) = 0 θ(0) = 0

The solution in MATLAB is given in the following file:

function yprime = ds2pt74(t,y)

m = 1;Fr = 3*t 2-1;Fth = cos(t*pi.6);

r = y(1); % radial positionth = y(2); % angular positionv = y(3); % radial velocityw = y(4); % angular velocity

yprime(1) = v;yprime(2) = w;yprime(3) = Fr/m + r*w 2;yprime(4) = (1/r)*(Fth/m-2*v*w);

return

152

The Mathcad solution follows:

m 1:=

ddr r dθ, t,( ) r dθ2

⋅1

m3 t

2⋅ 1−( )⋅+:=

ddθ dr r, dθ, t,( ) 2− dr⋅ dθ⋅

r

1

m r⋅cos

π t⋅

6

⋅−:=

i 0 2000..:= ∆t 0.001:= ti

i ∆t⋅:=

dr0

r0

dθ0

θ0

0

2

0

0

:=

dri 1+

ri 1+

dθi 1+

θ i 1+

dri

ddr ri

dθi, ti

,( )∆t⋅+

ri

dri∆t⋅+

dθi ddθ dri

ri

, dθi, ti

,( )∆t⋅+

θ i dθi ∆t⋅+

:=

xi

ri

cos θ i( )⋅:= yi

ri

sin θ i( )⋅:=

1.5 2 2.5 3 3.5

2

0

y i

xi

2.75 The force acting on m along r is:

Differentiating the given value of θ yields:

Fr = −k(r − 0.2)er.

θ = 0.1t2

θ = 0.2t

θ = 0.2

Using equation 2.39 yields:

m(r − rθ2) = −k(r − 0.2)

m(rθ + 2rθ) = 0

The governing differential equations and initial conditions are:

θ = −24rθ

θ = −2rrθ

k = 400N/m m = 4kg

θ(0) = 0 r(0) = 0 r(0) = 0.2 θ(0) = 0

153

The MATLAB code is:

function yprime = ds2pt75(t,y)

m = 4;k = 40;L = 0.2;

th = 0.1*t 2;thdot = 0.2*t;

yprime(1) = y(2);yprime(2) = y(1)*thdot 2-(k/m)*(y(1)-L);

return

The Mathcad solution is:

i . .0 10000

∆t 0.001k 40

m 4

t .i ∆t

θ .0.1 t 2

dθ .0.2 t

ddr ,r dθ .r dθ2 .

k

mr 0.2

dr0

r0

0

0.2

dri 1

ri 1

dri

.ddr ,ri

dθi

∆t

ri

.dri

∆t

0

30

6090

120

150

180

210

240270

300

330

0.3

0.2

0.1

0

Trajectory Plot0.33073

0.2ri

θi

Starting point

i i

ii

i

154

2.76 Solution:

θ

µFs

f = - k mg(k v___

v ||)

FIGURE S2.76

v = (rex + rθeθ). The unit vector along v for computing f is:

v

|v| = rex+rθeθ√r2+r2θ2

Equation (2.39) from the book yields:

m(r − rθ2) = −k(r − 0.2) − µrmg r√(r2+r2θ2)

or

r = rθ2 − km

(r − 0.2) − µk · g · r√r2+r2θ2

The MATLAB code is:

function yprimem = d s2pt75(t,y) m = 4; k = 20; l = 0.2; g = 9.81; mu = 0.2;th = 0.1*t 2; thdot = 0.2*t; v = sqrt(y(2) 2+(y(1)*thdot) 2);

if v==0,yprime(1) = y(2);yprime(2) = y(1)*thdot 2-(k/m)*(y(1)-1);elseyprime(1) = y(2);yprime(2) = y(1)*thdot 2-(k/m)*(y(1)-1)-mu*g*y(2)/v;end

return

155

The Mathcad solution is:

k 20

m 4

µ 0.2g 9.81i . .0 10000

∆t 0.001

t .i ∆t

θ t .0.1 t2

dθ t .0.2 t

ddr ,,,r dr dθ t .r dθ .k

mr 0.2 ..µ g

dr

dr 2 .r dθ2

dr0

r0

0

0.2

dri 1

ri 1

dri

.ddr ,,,ri

dri

dθ ti ti

∆t

ri

.dri

∆t

0

30

60

90

120

150

180

210

240

270

300

330

0.4

0.2

0

Particle movement 0-10 s0.515584

0.199999ri

θ ti

Starting pointof the particlet=0 s

i

2

2.77 We are given m = 4 kg, k = 650 N/m and θ = 1.5 rad/s

Differentiate:

r(θ) = 0.200(2 − cos θ)

r(θ) = 0.200(sin θ)θ

r(θ) = 0.200[(cos θ)θ2 + (sin θ)θ]

156

The equation of motion in the radial direction is:

Nr − k(r − 0.100) = m(r − rθ2), where θ = 1.5. Thus:

Nr = k[0.200(2 − cos θ) − 0.100] + m[0.200 cos θ(1.5)2 − 0.200(2 − cos θ)(1.5)2]

= 650(0.300 − 0.200 cos θ) + 4[0.400 cos θ(1.5)2 − 0.200(2)(1.5)2]

Thus

Nr = 191.4 − 126.4 cos θ.

2.78 We are given θ = 2 − cos θ, so that

θ = sin θθ = sin θ(2 − cos θ)

Thus

Nr = k(r − 0.100) + m(r − rθ2)

r(t) = 0.200[cos θ(2 − cos θ)2 + sin2 θ(2 − cos θ)]

rθ2 = 0.200(2 − cos θ)(2 − cos θ)2

The following MATLAB code computes the solution

function Nr = ds2pt78(theta)

thdot = 2-cos(theta);thddot = sin(theta).*thdot;

r = 0.2*(2-cos(theta));rdot = 0.2*sin(theta).*thdot;rddot = 0.2*sin(theta).*thddot + 0.2*cos(theta).*thdot. 2;

k = 650; m = 4; l = 0.1;

Nr = m*(rddot-r.*thdot. 2)+k*(r-l);

return

The following computes the solution in Mathcad:

157

θ 0π

48, 4 π⋅..:=

Fs θ( ) 650 0.2 2 cos θ( )−( )⋅ 0.1− ⋅:=

ddr θ( ) 0.2 cos θ( ) 2 cos θ( )−( )2⋅ sin θ( )( )2 2 cos θ( )−( )⋅+ ⋅:=

rωω θ( ) 0.2 2 cos θ( )−( )3⋅:=

Nr θ( ) Fs θ( ) 4 ddr θ( ) rωω θ( )−( )⋅+:=

0 200 400 600 8000

100

200

300

Nr θ( )

θ

deg

2.79 Following the solution of sample Problem 2.17 the following codes are used tonumerically adjust the mass until the desired response results. In MATLAB,the code is:

function yprime = ds2pt79(t,y)

m = 10.6;k = 500;r0 = 0.3;g = 9.81;

r = y(1);theta = y(2);rdot = y(3);thdot = y(4);

% velocitiesyprime(1,:) = rdot;yprime(2,:) = thdot;

% accelerationsyprime(3,:) = r*thdot 2-(k/m)*(r-r0)+g*cos(theta);yprime(4,:) = (-2*rdot*thdot-g*sin(theta))/r;

return

In Mathcad the code is:

158

k 1600

g 9.81N 2400

∆t 0.0005i . .0 N

v0

r0

ω0

θ

0

0.3

0

.30 deg

m:= 3.6 kg

The mass was adjusted by trial and error to produce the required path of motion.

a ,,,v r ω θ .r ω2 .g cos θ .

k

mr 0.3

α ,,,v r ω θ .1

r.g sin θ ..2 v ω

vi 1

ri 1

ωi 1

θi 1

vi

.a ,,,vi

ri

ωi

θi

∆t

ri

.vi

∆t

ωi

.α ,,,vi

r

i

ωi

θi

∆t

θ .ωi

∆t

xi

.ri

sin θ

yi

0.3 .ri

cos θ

-0.05 0.1 0.15

0

0.05Pendulum Trajectory

horizontal position (m)

yi

xi

0

i

i

i

0.05-0.1-0.15

-0.05

2.80 Solution:

159

mg

θk(r-L)

θ

FIGURE S2.80

From the free body diagram, the equations of motion can be written:

−k(r − L) + mg cos θ = mr − mrθ2

r = rθ2 + g cos θ − km

(r − L)

and:

−mg sin θ = 2mrθ + mrθ

θ = −grsin θ − 2 r

rθ

The MATLAB code to solve the problem numerically is:

function yprime = ds2pt80(t,y)m = 5;r = y(1);th = y(2);rdot = y(3);thdot = y(4);Fr = 1-sin(th);Fth = -(2+t-t 2);yprime(1) = rdot;yprime(2) = thdot;yprime(3) = r*thdot 2+Fr/m;yprime(4) = (-2*rdot*thdot+Fth/m)/r;return

The numerical solution in Mathcad is given on the following page. Note that thesolution (pendulum angle versus time) to Sample Problem 2.13 is also computedand plotted (in blue). The two solutions differ when k = 60 but are in goodagreement when k = 600. The length of the pendulum is also plotted as afunction of time showing the degree to which the mass is oscillating on thespring (note that the oscillations are very small when k = 600, i.e. in that casethe pendulum is behaving approximately as if the spring were a rigid bar orstring).

160

m 2:= L 2:= g 9.81:=

ddr r dθ, θ, k,( ) r dθ2

⋅ g cos θ( )⋅+k

mr L−( )⋅−:=

ddθ dr r, dθ, θ,( ) g−

rsin θ( )⋅ 2

dr dθ⋅

r⋅−:=

i 0 60000..:= ∆t 0.0001:= ti

i ∆t⋅:=

dr0

r0

dθ0

θ0

0

L

0

30 deg⋅

:=

dri 1+

ri 1+

dθi 1+

θ i 1+

dri

ddr ri

dθi, θ i, 60,( )∆t⋅+

ri

dri∆t⋅+

dθi ddθ dri

ri

, dθi, θ i,( )∆t⋅+

θ i dθi ∆t⋅+

:=

dβ0

β0

0

30 deg⋅

:=dβi 1+

βi 1+

dβig

Lsin βi( )⋅ ∆t⋅−

βi dβi ∆t⋅+

:=

0 2 4 650

0

50

θi

deg

β i

deg

ti

0 2 4 62

2.5

3

ri

ti

dr0

r0

dθ0

θ0

0

L

0

30 deg⋅

:=

dri 1+

ri 1+

dθi 1+

θ i 1+

dri

ddr ri

dθi, θ i, 600,( )∆t⋅+

ri

dri∆t⋅+

dθi ddθ dri

ri

, dθi, θ i,( )∆t⋅+

θ i dθi ∆t⋅+

:=

0 2 4 650

0

50

θi

deg

β i

deg

ti

0 2 4 61.5

2

2.5

3

ri

ti

161

2.81 The free-body diagram is:

R

mgN

θ

θ

FIGURE S2.81

Using equation (2.39) in the book the equations of motion are:

−N + mg cos θ = m(r − rθ2)

−mg sin θ = m(rθ + 2rθ)

Subject to: θ(0) = v0

R

we have that: θ = ωdωdθ

= − rR

sin θ, or:∫ ωv0/R ωdω =

∫ θ0 − g

Rsin θdθ

12(ω2 − v2

0

R2 ) = rR(cos θ − 1), thus: ω2 = 2g

R(cos θ − 1) +

v2

0

R2

The normal force will be a minimum at θ = 180◦

N = −mg + mRω2

Therefore ω2 ≥ rR

= 2gR

(−2) +v2

0

R2

v0 =√

5gR

2.82 The free-body diagram is:

mg

N

θ

f

FIGURE S2.82

From the diagram and equation (2.39) in the book:

−N + mg cos θ = −mRθ2

162

−mg sin θ − µk |N | θ|θ| = mRθ

Solving for the normal force:

N = mg cos θ + mRθ2

And substituting this into the equation for angular acceleration:

θ = − gR

sin θ − µk

R

∣

∣

∣g cos θ + Rθ2∣

∣

∣

θ|θ|

The computer solution in Mathcad is (independent of mass):

R 2:= µk 0.2:= v0 5:= g 9.81:=

α ω θ,( ) g−

Rsin θ( )⋅

µk

Rg cos θ( )⋅ R ω

2⋅+⋅

ω

ω⋅−:=

i 0 5000..:= ∆t 0.001:= ti

i ∆t⋅:=

ω0

θ0

v0

R

0

:=ωi 1+

θ i 1+

ωi α ωi θ i,( )∆t⋅+

θ i ωi ∆t⋅+

:=

0 2 4 650

0

50

100

θi

deg

ti

2.83 From the geometry in Figure S2.83a:

R/2

L

R

θ

β

FIGURE S2.83a

From the law of cosines:

L2 = R2 + R2

4− R2 cos(θ)

L = R2

√5 − 4 cos θ

163

And:R2

4= R2 + L2 − 2RL cos β

R2

4= R2 + R2

4(5 − 4 cos θ) − R2

√5 − 4 cos θ cos β

cos β = 2−cos θ√5−4 cos θ

From the law of sines:sin θL

= sin βR/2

sin β = sin θ√5−4 cos θ

The spring force is:

Fs = k(L − L0) = kR2

[√5 − 4 cos θ − 1

4

]

Components of the spring force are:

Fsn = −Fs cos β = −kR2

[

1 − 14√

5−4 cos θ

]

(2 − cos θ)

Fsθ = −Fs sin β = −kR2

[

1 − 14√

5−4 cos θ

]

sin θ

N

Fs

θe

> >

en

β

FIGURE S2.83b

From the free-body diagram in S2.83b, the equation of motion is:

mRd2θdt2

= −kR2

[

1 − 14√

5−4 cos θ

]

sin θ

d2θdt2

= − k2m

[

1 − 14√

5−4 cos θ

]

sin θ

164

2.84 Solution:

N

Fs

θe

> >

en

β

mg

θ

FIGURE S2.84

With the addition of the weight as shown in S2.84, the equation of motionbecomes:

mRd2θdt2

= −kR2

[

1 − 14√

5−4 cos θ

]

sin θ − mg cos θ

d2θdt2

= − k2m

[

1 − 14√

5−4 cos θ

]

sin θ − gR

cos θ

2.85 Solution:

N

Fs

θe

> >

en

βf

FIGURE S2.85

The friction force is:

f = −µkNθ|θ|

With the addition of the friction force as shown in S2.85, the equation of motionbecomes:

mRd2θdt2

= −kR2

[

1 − 14√

5−4 cos θ

]

sin θ − µkNθ|θ|

d2θdt2

= − k2m

[

1 − 14√

5−4 cos θ

]

sin θ − µkNmR

θ|θ|

where:

N = kR2

[

1 − 14√

5−4 cos θ

]

(2 − cos θ) − mRθ2

165

2.86 Solution:

N

Fs

θe

> >

en

β

mg

θf

FIGURE S2.86

With the addition of both the weight and the friction force as shown in S2.86,the equation of motion becomes:

mRd2θdt2

= −mg cos θ − kR2

[

1 − 14√

5−4 cos θ

]

sin θ − µkNθ|θ|

d2θdt2

= − gR

cos θ − k2m

[

1 − 14√

5−4 cos θ

]

sin θ − µkNmR

θ|θ|

where:

N = mg sin θ + kR2

[

1 − 14√

5−4 cos θ

]

(2 − cos θ) − mRθ2

A Mathcad code for integrating the equation of motion is as follows with somechoice of m, k, R, µk and v0 such the the motion damps out after two oscillations.There are infinitely many other combinations that would satisfy this criterion.

166

m 2:= k 70:= µ 0.3:= g 9.81:= R 1:=

f θ( ) 11

4 5 4 cos θ( )⋅−⋅−:=

N dθ θ,( ) kR

2⋅ f θ( )⋅ 2 cos θ( )−( )⋅ m R⋅ dθ

2⋅− m g⋅ sin θ( )⋅+:=

ddθ dθ θ,( ) k−

2 m⋅f θ( )⋅ sin θ( )⋅

g

Rcos θ( )⋅−

µ

m R⋅N dθ θ,( )⋅

dθ

dθ⋅−:=

i 0 40000..:= ∆t 0.0001:= ti

i ∆t⋅:=

dθ0

θ0

0

30 deg⋅

:=dθi 1+

θ i 1+

dθi ddθ dθi θ i,( )∆t⋅+

θ i dθi ∆t⋅+

:=

0 2 4100

50

0

50

θi

deg

ti

2.87 We are given:

M = 5.98 × 1024kg

r = 6380 × 103 m

G = 66.7 × 10−12 m3/kg · s2

F

m

FIGURE S2.87

GmMr2 = mv2

r

v =√

GMr

=√

5.98×66.7500+6380

× 103 = 7614 m/s

167

2.88 Solution:

Rω2m = GMmR2

ω =√

GMR3

The period τ is

τ = 2πω

= 2π√GM

R3

2

2.89 Solution:

τ = 23 hrs 56 min. = 86.16 ×103 s.

τ = 2π√GM

r3/20

r3/20 = τ

√GM

2π= 273.87 × 109 m

r0 = 42210 km.

r = r0 − R = 42210 − 6380

r = 35830 km

v =√

GMr0

= 3074 km/s.

2.90 Solution:

First, let us determine the velocity at final circular orbit of 700 km altitude:

R4 = 6380 + 700 = 7080 km

v4 =√

MGRA

= 7.506 × 103 m/s

R1 = 6380 + 400 = 6780 km

v1 =√

MGR1

= 7.670 × 103 m/s

From Eq. (2.67) for a conic section:1r

= A(1 + e cos θ)

For elliptic orbit r0 ⇒ θ = 0, rapogee ⇒ θ = 180

Therefore rapogee

r0= (1+e)

(1−e)

Solving for e: e = (ra/r0)−1(ra/r0)+1

ra

r0= 7080

6780= 1.044

e = 0.022

h is the angular momentum/unit mass = r0v0. From eq. 2.69

e = h2

GMr0− 1 =

r0v2

0

GM− 1

The velocity to go into elliptical orbit is

v2 =√

GMR1

(1 + e) = 7.754 × 103 m/s

168

Since angular momentum is concerned in elliptical orbit

v3ra = v2r0, or v3 = v267807080

= 7.411 × 103 m/s

Therefore the first boost v1 → v2 is 7.670 × 103 m/s or 7.754 × 103 m/s

The second boost v3 → v4 is

7.411 × 103 → 7.506 × 103 m/s.

2.91 From Problem 2.89

R4 = 42, 170 km

v4 = 3.075 × 103 m/s

From Problem 2.90

R1 = 6780 km,

v1 = 7.670 × 103 m/s

e = R4/R1−1R4/R1+1

= 0.723

v2 =√

GMR1

(1 + e) = 10.070 × 103 m/s

v3 = R1

R4v2 = 1.619 × 103 m/s

Therefore v1 → v2: 7.670 × 103 → 10.070 × 103 m/s

v3 → v4 : 1.619 × 103 → 3.075 × 103 m/s

2.92 For an elliptical orbit

e = 0.12 vp = 4000 m/s1r0

= GMh2 [1 + e], h = r0v0

r0 = GMv2

0

(1 + e), M = 5.98 × 1024 kg, G = 66.7 × 10−r m2/kg · s2

r0 = 5.98×66.7×1012

16×106 (1 + 0.12) = 27.92 × 106 m

Therefore k = 111.68 × 109 m2/s

At the apogee1ra

= GMh2 (1 − e)

1ra

= 5.98×66.7×10!2

111.683×1018 (0.88) = 2.8 × 10−8

ra = 35.7 × 106 m va = kra

= 3128 m/s

Altitude = ra − re = (35.7 − 6.378) × 106 = 29, 332 km

169

2.93 Solution:

For a parabolic arc: e = 11r

= GMk2 (1 + e cos θ)

r = 200, 000, 000, θ = 140◦

cos 140◦ = −0.766

Therefore h2 = 2 × 108 × 5.98 × 1024 × 66.7 × 10−12(1 − 0.766)

h = 136.6 × 109

d = hv

= 136.6×109

2.222×103

v

d

FIGURE S2.93

d = 23, 395 km

Re = 6380 km

Altitude = d − Re = 17, 015 km

2.94 Solution:

ra = 6380 × 103 + 392.62 × 103 = 6772.62 × 103 m

rp = 6380 × 103 + 385.42 × 103 = 6765.42 × 103 m

1rp

= GMh2 [1 + e], e = h2

GMr0

− 1

τ = 92.34 · 60 = 5.54 × 103 seconds

h = πτ(ra + rp)

√rarp = 5.196 × 1010

e = h2

GMrp− 1 = 5.108 × 10−4

2.95 Final velocity (Mir’s velocity at the perigee), using values from 2.94:

v4 =√

GMrp

(1 + e) = 7.687 × 103 m/s

R1 = 6380 + 100 = 6480 km

v1 =√

MGR1

=√

5.98×66.76.480

× 103 = 7.846 × 103 m/s

170

The Hohmann transfer-elliptic intercept orbit

R1 = 6480 km, rp = 6765 km

e2 =rpR1

−1rpR1

+1= 0.022

The velocity to go into this orbit is

v3 =√

GMR1

(1 + e2) = 7932 × 103 m/s

Conservation of momentum:

v3 = 7932 × 103 × R1

rp= 7598 × 103 m/s

v1 → v2 7.846 → 7.932 km/s

v3 → v4 7.598 → 7.687 km/s

2.96 Solution: v4 =√

MGR4

, R4 = 6380 + 150 = 6530 km

v4 = 7816 m/s

R1 = 6765 km v1 = 7687 m/s

rpra

FIGURE S2.96

e =4arp

−1rarp

+1, ra = 6765, rp = 6530 ra

rp= 1.036,

e = 0.018,

v2 to go into elliptic transfer orbit

v2 =√

GMR

(1 − e) = 7618 m/s

v3 = 7618 × ra

rp= 7892 m/s

Retrofiring

v1 → v2 7687 → 7618 m/s

v3 → v4 7892 → 7816 m/s

171

2.97 Solution:

mg

fw

Nw

N b

fb

β

FIGURE S2.97

w = −mgk

Nω = −NωeR

fω = +µNω(−eeθ cos β + k sin β)

Nb = Nb(eθ sin β + cos βk)

fb = µNb(−eθ cos β + k sin β)

The equations of motion are

−Nω = −mrθ2

−µNω cos β + Nb sin β − µNb cos β = mrθ

−mg + µNω sin β + Nb cos β + µNb sin β = mz

Constraint: a = a(+ cos βeθ − sin βk) − rθ2er

−µmrθ2 cos β + Nb(sin β − mu cos β) = ma cos β (1)

−mg + µmkθ2 sin β + Nb(cos β + µ sin β) = −ma sin β (2)

Mulp (1) by sin β and (2) by cosβ and add.

−mg cos β + Nb(cos2 β − µ sin β cos β + sin2 β − µ sin β cos β) = 0

Therefore Nb = mg cos β1−2µ sinβ cos β

a = 1m

(−µmrθ2 + mg(sin β−µ cos β)1−2µ sinβ cos β

)

Therefore

θ = α = 1mr

[

−µmrθ2 cos β + mg cos β(sin β−µ cos β)1−2µ sinβ cos β

]

z = 1m

[

−mg + µmrθ2 sin β + mg cos β(cos β+µ sin β)1−2µ sinβ cos β

]

z = +µrθ2 sin β − g sinβ(sin β−µ cos β)1−2µ sinβ cos β

The Mathcad solution:

172

i . .0 5000

∆t 0.001

t .i ∆t

g 9.81

β .30 deg

µ 0.1r 0.5

α ω .1

r....µ r ω ω cos β ..g cos β

sin β .µ cos β

1 ...2 µ sin β cos β

a ω ....µ r ω ω sin β ..g sin βsin β .µ cos β

1 ...2 µ sin β cos β

0 1 2 3 4 55

0

5Velocity in z direction

time s

vi

.r ωi

ti

0 2 4 60

20

40

θi

ti

vi 1

zi 1

ωi 1

θi 1

vi

.a ωi

∆t

zi

.vi

∆t

ωi

.α ωi

∆t

θ .ωi

∆ti

v0

z0

ω0

θ

0

0

0

00

i

173

2.98 We are given:

vθ = rθ, r = 1.5, h = 50, vθ0= 20

1.5

a = (r − rθ2)er + (2rθ + rθ)eθ + zk

mg

N

FIGURE S2.98

F = −N ex − mgk

r = r = 0

N = mrθ2

0 = rθ

−mg = mz

θ = 0, θ = θ0, θ = θ0t

z = −g, z = −gt, z = −12ft2 + z0

The following MATLAB code plots this:

clear allformat short eformat compact

t = linspace(0,2);

g = 9.81;

v0 = 20;R = 1.5;w0 = v0/R;

theta = w0*t;z = -(g/2)*t. 2;r = R*ones(size(t));

x = r.*cos(theta);y = r.*sin(theta);

174

plot3(x,y,z)

The Mathcad code is:

i . .0 200

t

i

i

100

ω20

1.5

x .1.5 cos .ω ti

y .1.5 sin .ω ti

zi

..1

29.81 t

i

2 50

3-D trajectory

1 0 110

1

35

40

45

50

,,x y z

175

2.99 Solution:

v =

0r · ωvz

f = −µNv/|v|f = −µmrθ2√

r2ω2+v2g

(rωeθ + vzk)

mrθ = − µmrθ2√r2ω2+v2

3

rθ, z = −g − µrθ2vz√r2ω2+v2

z

These are solved numerically by the following MATLAB code:

function yprime = ds2pt99(t,y)

R = 1.5;g = 9.81;mu = 0.2;

th = y(1);z = y(2);thdot = y(3);zdot = y(4);

v = sqrt((R*thdot) 2+zdot 2);vprime(1) = thdot;yprime(2) = zdot;yprime(3) = -mu*R*thdot 3/v;yprime(4) = -mu*R*thdot 2*zdot/v-g;

return

176

The Mathcad code follows:

i . .0 390

∆t 0.01

t .i ∆t g 9.81 µ 0.2 r 1.5

α ,vz ω..µ r ω

3

.r ω2

vz2

a ,vz ω g...µ r ω

2vz

.r ω2

vz2

vz0

z0

ω0

θ

0

50

20

1.5

0

vzi 1

zi 1

ωi 1

θi 1

vzi

.a ,vzi

ωi

∆t

zi

.vzi

∆t

ωi

.α ,vz

i

ωi

∆t

θ .ωi

∆t

n . .0 391

x .r cos θn

y .r sin θn

zn

zn

3-D Trajectory

1 0 1101

0

20

40

,,x y z

177

2.100 This follows 2.99 and the codes are the same with different initial condition.The Mathcad version follows:

i . .0 390

∆t 0.01

t .i ∆t g 9.81 µ 0.2 r 1.5

α ,vz ω..µ r ω

3

.r ω2

vz2

a ,vz ω g...µ r ω

2vz

.r ω2

vz2

vz0

z0

ω0

θ

20

30

20

1.5

0

vzi 1

zi 1

ωi 1

θi 1

vzi

.a ,vzi

ωi

∆t

zi

.vzi

∆t

ωi

.α ,vz i ωi

∆t

θ .ωi

∆t

n . .0 391

x .r cos θn

y .r sin θn

zn

zn

0

i

n

n

178

2.101 We have vθ = v0 cos θ and vz = v0 sin β

mg

N

yz

x

voβ

θ

FIGURE S2.101

N = −N eR

W = mg(cos θer − sin θeθ)

mz = 0 (1)

−mrθ2 = −N + mg cos θ (2)

mrθ = −mg sin θ (3)

Eq. (3) may be written

ω dωdθ

= −g/r sin θ

Integrating yields:ω2

2|ωω0

= g/r cos θ|θ0ω2 − ω2

0 = 2gr(cos θ − 1) (4)

From Eq. 2, minimum velocity can be determined when θ = 180◦ and N = 0.The angular velocity ω2

180 = g/r. From (4)

g/r − ω20 = −7g

rTherefore ω0 =

√

5gr

vθ0=

√4gr = v0 cos β, v0 =

√5gR

cos β.

2.102 From the previous problem the forces are

N = −N er

W − mg(cos θer − sin θeθ)

f = −µN√(rω)2+vz2

(rωeθ + vzez)

rω0 = v0 cos β

vz0 = v0 sin β

−N + mg cos θ = −mrω2

179

mz = −µNvz/√

(rω)2 + vz2

mrθ = −mg sin θ − µNrω/√

(rω)2 + vz2

From 2.101 ω =√

g/r when θ = 180◦

N = mg cos θ + mrω2

Therefore z = −µ(g cos θ+rω2)vz√(rω)2+vz2

θ = −rrsin θ − µ(g cos θ+rω2)ω√

(rω)2+(vz)2

The minimum initial velocity of 10.6 m/s was determined by trial and errorusing the following codes. The motion damps out after 4 seconds and theparticle comes to rest at 360 degrees. The MATLAB code is:

function yprime = ds2pt102(t,y)

m = 3;R = 0.6;g = 9.81;mu = 0.2;

th = y(1);z = y(2);thdot = y(3);zdot = y(4);

v = sqrt((R*thdot) 2+zdot 2);N = m*g*cos(th)+m*R*thdot 2;

yprime(1) = thdot;yprime(2) = zdot;yprime(3) = (-m*g*sin(th)-mu*N*R*thdot/v)/m/R;yprime(4) = (-mu*N*zdot/v)/m;

return

180

The Mathcad code is:

I:= 0 ..5000

∆t 0.001

t .i ∆t

µ 0.2 β .50 deg g 9.81 v0 10.6 r 0.6

V ,vz ω .r ω2

vz 2

a ,,vz ω θ ..µ .g cos θ .r ω2 vz

V ,vz ω

α ,,vz ω θ .g

rsin θ ..µ .g cos θ .r ω

2 ω

V ,vz ω

vz0

z0

ω0

θ

.v0 sin β

0

.v0cos β

r

0

vzi 1

zi 1

ωi 1

θi 1

vzi

.a ,,vzi

ωi

θi

∆t

zi

.vz

i

∆t

ωi

.α ,,vz ωi

θ ∆t

θi

.ωi

∆t

ω ming

r

=ω min 4.044

0

i

i

i

181

0 100 200 300 400 50010

0

10

20

angular position degrees

ωi

θ i

deg

0 1 2 3 4 50

200

400

600Angle vs time

time s

θi

deg

ti

The minimum initial velocity of 10.6 m/s was determined by trial and error. The motion damps out after 4 seconds and the particle comes to rest at 360 degrees.

182

2.103 The forces are:

3

β

x

y

N

mg

θ

FIGURE S2.103

N = −N er

W = mg(cos β sin θer + cos β cos θeθ + sin βk)

The equations of motion are:

−N + mg cos β sin θ = −mrθ2

mg cos β cos θ = mrθ

mg sin β = mz

The data is:m = 2 kg θ(0) = 0r = 1 mβ = 20◦

The z equation can be quickly integrated:

z = g sin βz = g(sin β)t

z = g(sin β)t2/2

The MATLAB code is:

function yprime = ds2pt103(t,y)

g = 9.81;R = 1;beta = 20*pi/180;

yprime(1) = y(2);yprime(2) = (g/R)*cos(beta)*cos(y(1));

return

183

The Mathcad code is:

i . .0 2500

∆t 0.001

t .i ∆t

β .20 degr 1g 9.81

α θ ..g

rcos β cos θ

ω0

θ

0

0

ωi 1

θi 1

ωi

.α θi

∆t

θ .ωi

∆t

n . .0 2500

zn

..g sin βtn

2

2

0 0.5 1 1.5 2 2.5100

0

100

200Angular position vs time

time s

θi

deg

ti

0 0.5 1 1.5 2 2.50

5

10

15Position down sluiceway

time s

z i

ti

0

i

i

184

2.104 We are given: µk = 0.3

Friction acts in a direction opposite to the velocity

v = rωeθ + vzk

N = −N er

W = mg(cos β sin θer + cos β cos θeθ + sin βk)

f = −Nµ v|v|

The equations of motion are:

−N + mg cos β sin θ = −mrω2

mg cos β cos θ − µN rω√(rω)2+vz2

= mrθ

mg sin β − µN vz√(rω)2+vz2

= mz

Therefore N = m(g cos β sin θ + rω2) and

α(vz, ω, θ) = 1r

[

g cos β cos θ − µNm

rω√(rω)2+vz2

]

a(vz, ω, θ) = g sin β − µNm

vz√(rω)2+vz2

.

The MATLAB code is:

function yprime = ds2pt103(t,y)

m = 2;g = 9.81;R = 1;mu = 0.3;beta = 20*pi/180;

th = y(1);z = y(2);thdot = y(3);zdot = y(4);

v = sqrt((R*thdot) 2 + zdot 2);N = m*g*cos(beta)*sin(th)+m*R*thdot 2;

yprime(1) = thdot;yprime(2) = zdot;yprime(3) = (m*g*cos(beta)*cos(th)-mu*N*R*thdot/v)/m/R;yprime(4) = (m*g*sin(beta)-mu*n*zdot/v)/m;

return

185

The Mathcad code is:

i . .0 6000

∆t 0.001

t .i ∆t

β .20 deg r 1 g 9.81

N ,ω θ ..g cos β sin θ .r ω2

α ,,vz ω θ .1

r..g cos β cos θ ..µ N ,ω θ

.r ω

.r ω2

vz2

a ,,vz ω θ .g sin β ..µ N ,ω θvz

.r ω2

vz2

vz0

z0

ω0

θ

0

0

0

0

vzi 1

zi 1

ωi 1

θi 1

vzi

.a ,,vzi

ωi

θi

∆t

zi

.vz ∆t

ωi

.α ,,vz ωi

θi

∆t

θi

.ωi

∆t0

i

i

i

0 1 2 3 4 5 60

100

200

time s

θi

deg

ti

0 1 2 3 4 5 60

10

20Position along Sluiceway

time s

z i

ti

186

2.105 We have m = 2 kg, R = 1.5 m, v(0) = 10eθ, φ(0) = 90

θ

φ

N

mg

FIGURE S2.105

The forces are:

N = −N eR

W = mg(sin φeφ − cos φeR)

The coordinate accelerations are

aR = R − Rφ2 − R sin2 φθ2

aφ = Rφ + 2Rφ − R sin φ cosφθ2

aθ = R sin φθ + 2Rθ sin φ2Rφθ cos φ

Therefore −N − mg cos φ = −mR(φ2 + sin2φθ2)

mg sin φ = mR(φ − sin φ cosφθ2)

0 = R sin θθ + 2Rφθ cos φ

From sample problem 2.22, the last equation

θ = sin φ0vθ0

R sin2 φ= 10

R sin2 φ

φ = gR

sin φ +cos φv2

θ0sin2 φ0

R2 sin3 φ

The MATLAB code is:

function yprime = ds2pt105(t,y)

g = 9.81;R = 1.5;

ph = y(1);th = y(2);phdot = y(3);

yprime(1) = y(3);yprime(2) = 10/(R*sin(ph) 2);yprime(3) = -(g/R)*sin(ph) + 100*cos(ph)/(R 2*sin(ph) 3);

return

187

The Mathcad code is (first define R and g in the code):

∆t 0.001

t .i ∆t

ddφ φ .gsin φ

R

.100 cos φ

.R 2 sin φ3

dφ0

φ0

0

.90 deg

dφi 1

φi 1

dφi

.ddφ φi

∆t

φi

.dφi

∆t

dθ φ10

.R sin φ2

θ 0

θi 1

θ .dθ φi

∆t

x ..R sin φi

cos θ

y ..R sin φi

sin θ

zi

.R cos φi

=φ0

1.571

3-D Trajectory

10

11

0

1

0.40.3

0.20.1

0

,,x y z

i

i

i

i

0

i

i

188

2.106 We are given: R = 1.5, θ0 = 101.5

, θ0 = 0, φ0 = 90◦

v = Rφeφ + R sin φθeθ

N = −N eR

W = mg(sin φeφ − cos φer)

f = −µN v

|v|

−N − mg cos φ = −mR(φ2 + sin2 φθ2)

mg sin φ − µN Rφ|v| = mR(φ − sin φ cos φθ2)

−µN R sinφθ|v| = mR(sin φθ + 2φθ cos φ)

The mass drops out.

N/m = R(φ2 + sin2 φθ2) − g cos φ

Therefore

φ = rR

sin φ − µR(φ2 + sin2 φθ2) φ|v| + sin φ cos φθ2

θ = 1sin φ

[

−µNm

sinφθv

− 2φθ cos φ]

The MATLAB code is:

function yprime = ds2pt105(t,y)

g = 9.81;R = 1.5;mu = 0.3;

ph = y(1);th = y(2);phdot = y(3);thdot = y(4);

v = R*sqrt(phdot 2+sin(ph) 2*thdot 2);N = g*cos(ph)+v 2/R;

yprime(1) = y(3);yprime(2) = y(4);yprime(3) = - (g/R)*sin(ph) -mu*Nphdot/v+sin(ph)*cos(ph)*thdot 2;yprime(4) = -mu*N*thdot/v-2*phdot*thdot/tan(ph);

return

189

The Mathcad code is:

R 1.5 g 9.81 µ 0.3i . .0 4000

∆t 0.001

t .i ∆t

N ,,dφ φ dθ .R dφ2 .sin φ

2dθ .g cos φ

v ,,dφ φ dθ .R dφ2 ..R sin φ dθ

2

ddφ ,,dφ φ dθ .g

Rsin φ ..µ N ,,dφ φ dθ

dφ

v ,,dφ φ dθ..sin φ cos φ dθ

2

ddθ ,,dφ φ dθ .1

sin φ...µ N ,,dφ φ dθ sin φ

dθ

v ,,dφ φ dθ...2 dφ dθ cos φ

dφ0

φ0

dθ0

θ

0

.90 deg

10

1.5

0

dφi 1

φi 1

dθi 1

θi 1

dφi

.ddφ ,,dφi

φi

dθi

∆t

φi

.dφi

∆t

dθ .ddθ ,,dφi

φi

dθi

∆t

θ .dθi

∆t

x ..R sin φi

cos θ

y ..R sin φi

sin θ

zi

.R cos φi

i

i

i

0

i

i

i

i

2.107 Solution:

z

yx

xz

y

_ R

N

mg

ψ

ψ

e

w_

ϕ

ϕ

FIGURE S2.107

190

N = −N eφ

W = mg(− cos φee + sin φeφ)

f = −µN v

|v|

aR = R − Rφ2 − R sin2 φθ2

aφ = Rφ + 2Rφ − R sin φ cos φθ2

aθ = R sin φθ + 2Rθ sin φ + 2Rφθ cos φ

φ = const

v = ReR + R sin φθeθ

−mg cos φ − µN Rv

= m(R − R sin2 φθ2)

−N + mg sin φ = −mR sin φ cos φθ2

−µN R sinφθv

= m(R sin φθ + 2Rθ sin φ)

N/m(dθ, R) = g sin φ + R sin φ cos φθ2

v(dR, R, dθ) =√

dR2 + (R sin φdθ)2

R = −g cos φ + R sin2 φθ2 − µNm

R/v

θ = −µ vm

θv− 2Rθ

R

The MATLAB code is:

function yprime = ds2pt107(t,y)

g = 9.81;mu = 0;phi = pi/6;

r = y(1);th = y(2);rdot = y(3);thdot = y(4);

v = sqrt(rdot 2+(r*sin(phi)*thdot) 2);N = g*sin(phi)+r*sin(phi)*cos(phi)*thdot 2;

yprime(1) = rdot;yprime(2) = thdot;yprime(3) = -mu*N*rdot /v - g*cos(phi)+r*sin(phi) 2*thdot 2;yprime(4) = -mu*N*thdot/v - 2*rdot*thdot/r;

return

191

The Mathcad code is:

i . .0 6000

∆t 0.001

t .i ∆t

φ .30 deg

g 9.81 µ 0.0

R02

cos φN ,dθ R .g sin φ ...R sin φ cos φ dθ

2

v ,,dθ dR R dR2 ..R sin φ dθ2

ddR ,,dθ dR R .g cos φ ..R sin φ2

dθ ..µ N ,dθ RdR

v ,,dθ dR R

ddθ ,,dθ dR R ..µ N ,dθ Rdθ

v ,,dθ dR R.2

.dR dθ

R

dR 0

R0

dθ

0θ

0

R0

10

.R0 sin φ

0

dRi 1

Ri 1

dθi 1

θi 1

dR .ddR ,,dθ

i

dR Ri

∆t

R

i

.dRi

∆t

dθ

i

.ddθ ,,dθ

i

dR

i

Ri

∆t

θ .dθi ∆t

x ..R sin φ cos θ

y ..R sin φ sin θ

zi

.Ri

cos θ

i

0

i

i

ii

i

i

i

i

i

ii

iii

Path of Motion

202

0

5

0

,,x y z

192

2.108 Use the analysis and codes of 2.107 with a new friction coefficient. The codesare essentially the same, so only the Mathcad plot is shown.

Path of Motion

05050

0

50

100

0

100

,,x y z

2.109 Solution:

mg

N

FIGURE S2.109

N = N eR

W = mg(− cos φeR + sin φeφ)

V = Rφeφ + R sin φθeθ

f = −µN v

|v|

N − mg cos φ = −m(Rφ2 + R sin2 φθ2)

mg sin φ − µN Rφv

= m(Rφ − R sin φ cosφθ2)

−µN R sinφθv

= m(R sin φθ + 2Rφθ cos φ)

193

v(dφ, φ, dθ) = R ·√

dφ2 + (sin(φ)dθ)2

N/m = g cos φ − R(φ2 + sin2 φθ2)

φ = gR

sin φ − µNm

φv

+ sin φ cosφθ2

θ = −µNm

θv− 2φθ cos φ

sinφ

The MATLAB code is:

function yprime = ds2pt107(t,y)

g = 9.81;mu = 0;r = 6;

ph = y(1);th = y(2);phdot = y(3);thdot = y(4);

v = r*sqrt(phdot 2+(sin(ph)*thdot) 2);N = g*cos(ph) -v 2/r;

yprime(1) = phdot;yprime(2) = thdot;yprime(3) = -mu*N*phdot/v + (g/4)*sin(ph) +sin(ph)*cos(ph)*thdot 2;yprime(4) = -mu*N*thdot/v -2*phdot*thdot/tan(ph);

return

194

The Mathcad code is:

i . .0 900

∆t 0.001

t .i ∆t

R 6

g 9.81

µ 0.0

v ,,dφ φ dθ .R dφ2 .sin φ dθ

2

N ,,dφ φ dθ .g cos φ .R dφ2 .sin φ dθ

2

ddφ ,,dφ φ dθ .g

Rsin φ ..µ N ,,dφ φ dθ

dθ

v ,,dφ φ dθ..sin φ cos φ dθ

2

ddθ ,,dφ φ dθ ..µ N ,,dφ φ dθdθ

v ,,dφ φ dθ...2 dφ dθ

cos φ

sin φ

dφ0

φ0

dθ

0θ

0

0.5

6

.10 deg

4

.6 sin .10 deg

0

dφi 1

φi 1

dθi 1

θi 1

dφi

.ddφ ,,dφi

φi

dθi

∆t

φi

.dφi

∆t

dθ i.ddθ ,,dφ

iφ

idθ

i∆t

θ .dθi

∆t

i

i

0 200 400 600 800 10005

0

5

10

N ,,dφi

φi

dθi

in 866

=N ,,dφn

φn

dθ 3.95 103

=φ

866

deg41.541

=θ

deg75.735

The particle will fall off the surface when the normal force goes to zero and the positioncoordinates are shown.

n

866

195

2.110 Use the same codes and analysis as in 2.109. The MATLAB code becomes:

clear allformat short 3format compact

g = 9.81;r = 6;

ph0 = pi/18;th0 = 0;phdot0 = -0.5/r;thdot0 = 4/r/sin(ph0);

[t,Y] = ode45(’ds2pt109’,0,1,[ph0 ; th0; phdot0; thdot0]);

ph = Y(:.1);th = Y(:.2);phdot = Y(:.3);thdot = Y(:.4);

v = r*sqrt(phdot. 2+(sin(ph).*thdot). 2);N = g*cos(ph( - v. 2/r;

figure(1),plot(t,N)

196

The Mathcad code is:

i . .0 2000

∆t 0.001

t .i ∆t

R 6

g 9.81

µ 0.2

v ,,dφ φ dθ .R dφ2 .sin φ dθ

2

N ,,dφ φ dθ .g cos φ .R dφ2 .sin φ dθ

2

ddφ ,,dφ φ dθ .g

Rsin φ ..µ N ,,dφ φ dθ

dθ

v ,,dφ φ dθ..sin φ cos φ dθ

2

ddθ ,,dφ φ dθ ..µ N ,,dφ φ dθdθ

v ,,dφ φ dθ...2 dφ dθ

cos φ

sin φ

dφ0

φ0

dθ

0θ

0

0.5

6

.10 deg

4

.6 sin .10 deg

0

dφi 1

φi 1

dθi 1

θi 1

dφi

.ddφ ,,dφi

φi

dθi

∆t

φi

.dφi

∆t

dθ i.ddθ ,,dφ

iφ

idθ

i∆t

θ .dθi

∆ti

0 500 1000 1500 200010

0

10

N ,,dφi

φi

dθi

in 1816

=N ,,dφn

φn

dθ 0.011

=φ

n

deg51.168

=θ

n

deg113.121

The mass stays on the surface for a greater length of time when friction is present. If thefriction is too great, the particle will not leave the surface before it stops.

n

-

197

2.111 The data is:

y x

zmg

ϕ

θ

FIGURE S2.111

φ(0) = 30◦ θ(0) = 0 m = 3 kg R(0) = 2 vθ = 0.5, therefore θ(0) = 0.5.

Fs = −k(R − R0)eF

W = mg(cos φeR − sin φeφ)

The equation of motion become:

m[R − Rφ2 − R sin2 φθ2] = mg cos φ − k(R − 2)

m[Rφ + 2Rφ − sin φ cos φθ2] = −mg sin φ

m[R sin φθ + 2Rθ sin φ + 2Rφθ cos φ] = 0, or:

R = g cos− km

(R − 2) + Rφ2 + R sin2 φθ2

φ = − gR

sin φ − 2RφR

+ sin φ cos φθ2

θ = −2RθR

− 2φθ cos φsinφ

The last equation can be written1

R sin φddt

[R2 sin2 φθ] = 0

Therefore

R2 = sin2 φθ = h

h = R20 sin2 φ0θ0 = R0 sin φ0vθ0

For this case h = 2 sin 30◦(0.5) = 0.5

θ = hR2 sin2 φ

The differential equations may be written as

R = g cos φ − km

(R − 2) + Rφ2 + R sin2 φh2

R4 sin4 φ

R = g cos φ − km

(R − 2) + Rφ2 + h2

R3 sin2 φ

φ = − gR

sin φ − 2RφR

+ sin φ cos φh2

R4 sin4 φ

198

The MATLAB code is:

function yprime = ds2pt111(t,y)

m = 3;k = 200;g = 9.81;R = 2;v0 = 0.5;

h = v0*R*sin(pi/6);

r = y(1);ph = y(2);th = y(3);rdot = y(4);phdot = y(5);

thdot = h/r*sin(ph)) 2;

yprime(1) = rdot;yprime(2) = phdot;yprime(3) = thdot;yprime(4) = -(k/m)*(r-R)+g*cos(ph) + r*phdot + 4*sin(ph) 2*thdot 2;yprime(5) = -(g/r)*sin(ph) -2*(rdot/r)*phdot + sin(ph)*cos(ph)*thdot 2;

return

199

The Mathcad code is:

h 0.5 g 9.81 k 200 m 3

i . .0 5000

∆t 0.001

t .i ∆t

ddR ,,,dR R dφ φ .g cos φ .k

mR 2 .R dφ 2 h2

.R 3 sin φ2

ddφ ,,,dR R dφ φ .g

Rsin φ .2

.dR dφ

R

.cos φ h2

.R 4 sin φ3

dR0

R0

dφ0

φ0

0

2

0

.30 deg

dRi 1

Ri 1

dφi 1

φi 1

dR i.ddR ,,,dR

iR

i

dφi

φi

∆t

Ri

.dR

i

∆t

dφi

.ddφ ,,,dR

i

Ri dφi

φi

∆t

φi

.dφi

∆t

θ 0

θi 1

.h

.Ri

2 sin φi

2∆t θi

x ..R sin φi

cos θ yi

..R sin φi

sin θ zi

.Ri

cos φiiii i i

0

i . .0 5000

i

200

Path of motion

1

0

1

1

0.5

0

0.5

1

2

1

0

1

,,y x z

2 1 0 1 20.5

0

0.5

yi

xi

2 1 0 1 22.2

2

1.8

1.6

zi

xi

201

2.112 The MATLAB code is:

clear allformat short eformat compact

N = 100;

W = 3000;g = 32.2;

s = linspace(0,6000,N);v = 88;

th = s/2000 +exp(-s/1000);be = (pi/12)*sin(pi*s/3000);

th = 1/2000 -(1/100)*exp(-s/1000);be = (pi 2/36000)*cos(pi*s/3000);

Gvec = [ -sin(th).*cos(be).*th -cos(th).*sin(be.*be ;...cos(th).*cos(be).*th -sin(th).*sin(be.*be ;...cos(be).*be ];

for i = 1:N,mag Fn(i) = (W/g)*v 2*norm(Gvec(:,i));end

Figure (1),plot(s,mag Fn)xlabel(’distance along road (ft)’)ylabel(’normal force (lbf)’)grid

202

The Mathcad code is:

s . .,0 10 6000

W 6000

g 32.2

θ ss

2000e

s

1000

β s .π

12sin

.π s

3000

dθ s1

2000

1

1000e

s

1000

dβ s .π

36000cos

.π s

3000

Γ s

..sin θ s cos β s dθ s ..cos θ s sin β s dβ s

..cos θ s cos β s dθ s ..sin θ s sin β s dβ s

.cos β s dβ s

v s 88

F n s ..W

gv s 2 Γ s

0 1000 2000 3000 4000 5000 60000

500

1000

Position ft

F n s

s

203

2.113 Solution:

a = 2, 0 ≤ s ≤ 3000

v =√

2as

v(3000) = 109.5 ft/s (75 mph)

a = −1 3001 ≤ s ≤ 6000∫ v109.5 vdv =

∫ s3000 ads

v2

2− (109.5)2

2= −s + 3000

v =√

109.52 − 2(s − 3000)

The Mathcad code follows:

s . .,0 10 6000

W 6000

g 32.2

θ ss

2000e

s

1000

β s .π

12sin

.π s

3000

dθ s1

2000

1

1000e

s

1000

dβ s .π

36000cos

.π s

3000

Γ s

..sin θ s cos β s dθ s ..cos θ s sin β s dβ s

..cos θ s cos β s dθ s ..sin θ s sin β s dβ s

.cos β s dβ s

v s .Φ 3000 s .4 s .Φ s 3001 109.52 .2 s 3000

F n s ..W

gv s 2 Γ s

0 1000 2000 3000 4000 5000 60000

500

1000

1500

Position ft

F n s

s

204

2.114 Using the analysis of 2.113 the MATLAB code is:

clear allformat short eformat compact

N = 100;

W = 3000;g = 32.2;

s = linspace(0,6000,N)’;v = sqrt( 4*s.(s>=0) -6*(s-3000).*(s>=3000));a = 2*(s>=0) -3*(s>=3000);

t = s/2000 +exp(-s/1000);b = (pi/12)*sin(pi*s/3000);

t = 1/2000 - (1/1000)*exp(-s/1000);b = (pi 2/36000)*cos(pi*s/3000);

avec = [...a.*cos(t).*cos(b)+v. 2.*(-sin(t).*cos(b).*t -cos(t).*sin(b).*b ),...a.*sin(t).*cos(b)+v. 2.*( cos(t).*cos(b).*t -sin(t).*sin(b).*b ),...a.*sin(b)+v. 2.*cos(b).*b ];

mag a = sqrt(sum((avec.*avec)’)’);

figure(1),plot(s,mag a)xlabel(’distance along road (ft)’)ylabel(’acceleration (ft/sˆ2)’)grid

205

The Mathcad code is:

s . .,0 10 6000

W 6000

g 32.2

θ ss

2000e

s

1000

β s .π

12sin

.π s

3000

dθ s1

2000

1

1000e

s

1000

dβ s .π

36000cos

.π s

3000

Γ s

..sin θ s cos β s dθ s ..cos θ s sin β s dβ s

..cos θ s cos β s dθ s ..sin θ s sin β s dβ s

.cos β s dβ s

v s .Φ 3000 s .4 s .Φ s 3001 109.52 .2 s 3000

a n s ..1 v s 2Γ s

a t s .Φ 3000 s 2 Φ s 3001

a s a n s 2 a t s 2

0 1000 2000 3000 4000 5000 60002

4

6

8Acceleration Magnitude vs position

Position ft

a s

s

206

2.115 Solution:

fN

mg

FIGURE S2.115

F = −mgk + N n − µt = ma

n = Γ|Γ| f = −µN v

|v|

t = cos θ(s) cos β(s)i + sin θ(s) cos β(s)j + sin β(s)k]

N = mgk · n + m · v2

p

a(s) = −mgk·t−µNm

The MATLAB code is:

function yprime = ds2pt115(t,y)

m = 30;g = 9.81;mu = 0.0;

s = y(1);v = y(2);

th = pi*s/20;be = -(pi/3)*(1-(s/10). 2);

th = pi/20;be = 2*pi*s/300;

kvec = [ 0 ; 0 ; 1 ];

tvec = [...cos(th).*cos(be);...sin(th).*cos(be);...sin(be) ];

Gvec = [...-sin(th).*cos(be).*th -cos(th).*sin(be).*be ;...cos(th).*cos(be).*th -sin(th).*sin(be).*be ;...cos(be).*be ];

207

nvec = Gvec/norm(Gvec);

N = m*g*(kvec’*nvec) + v 2*norm(Gvec);

yprime(1) = y(2);yprime(2) = -g*(kvec’*tvec) - mu*N/m;

return

The Mathcad code is:

m 30

µ 0g 9.81

dθ sπ

20

θ s.π s

20

dβ s ..π

302

s

10

β s .π

31

s

10

2

Γ s

..sin θ s cos β s dθ s ..cos θ s sin β s dβ s

..cos θ s cos β s dθ s ..sin θ s sin β s dβ s

.cos β s dβ s

t s

.cos θ s cos β s

.sin θ s cos β s

sin β s

n sΓ s

Γ s

k

0

0

1

N ,v s ...m g k n s ..m v Γ s

a ,v s ..g k t s ..µ

mN ,v s

v

v

208

i . .0 1580

∆t 0.001

t .i ∆t

v0

s0

0

0

vi 1

si 1

v .a ,v

i

si

∆t

si

.v ∆t

=s1580

10.008

0 0.5 1 1.5 20

5

10

15Position on slide vs time

time s

s

i

ti

=v 10.987 m/s

This is the velocity at the bottom of the slide. Note the time at the bottom was found by trial and error and for this case was 1.58 s.

i

i i

1580

209

2.116 Use the analysis of 2.115. The MATLAB code is:

function yprime = ds2pt116(t,y)

m = 30;g = 9.81;mu = 0.2;

s = y(1);v = y(2);

th = pi*s/20;be = - (pi/3)*(1-(s/10). 2);

th = pi/20;be = 2*pi*s/300;

kvec = [ 0 ; 0 ; 1 ];

tvec = [...cos(th).*cos(be);...sin(t).*cos(be);...sin(be) ];

Gvec = [...-sin(th).*cos(be).*th -cos(th).*sin(be).*be ;...cos(th).*cos(be).*th -sin(th).*sin(be).*be ;...cos(be).*be [;

nvec = Gvec/norm(Gvec);

N = m*g*(kvec’*nvec) + v 2*norm(Gvec);

yprime(1) = y(2);yprime(2) = -g*(kvec’*tvec) - mu*N/m;

return

210

The Mathcad code is:

m 30

µ 0.2

g 9.81

dθ sπ

20

θ s.π s

20

dβ s ..π

302

s

10

β s .π

31

s

10

2

Γ s

..sin θ s cos β s dθ s ..cos θ s sin β s dβ s

..cos θ s cos β s dθ s ..sin θ s sin β s dβ s

.cos β s dβ s

t s

.cos θ s cos β s

.sin θ s cos β s

sin β s

n sΓ s

Γ s

k

0

0

1

N ,v s ...m g k n s ..m v Γ s

a ,v s ..g k t s ..µ

mN ,v s

v

v

211

i . .0 1712

∆t 0.001

t .i ∆t

v0

s0

0

0

vi 1

si 1

v i.a ,v

i

si

∆t

si

.v ∆t

=s1712

10.001

0 0.5 1 1.5 20

5

10

15Position on slide vs time

time s

si

ti

=v 7.9661712

i

i

212